About the HELM Projectqub.ac.uk/helm/HELM_2008/pages/...Vector_Calculus.pdf · Integral Vector Calculus 29.1 Line Integrals 2 29.2 Surface and Volume Integrals 34 29.3 Integral Vector

About the HELM ProjectHELM (Helping Engineers Learn Mathematics) materials were the outcome of a three-year curriculumdevelopment project undertaken by a consortium of five English universities led by Loughborough University,funded by the Higher Education Funding Council for England under the Fund for the Development ofTeaching and Learning for the period October 2002 September 2005.HELM aims to enhance the mathematical education of engineering undergraduates through a range offlexible learning resources in the form of Workbooks and web-delivered interactive segments.HELM supports two CAA regimes: an integrated web-delivered implementation and a CD-based version.HELM learning resources have been produced primarily by teams of writers at six universities:Hull, Loughborough, Manchester, Newcastle, Reading, Sunderland.HELM gratefully acknowledges the valuable support of colleagues at the following universities and col-leges involved in the critical reading, trialling, enhancement and revision of the learning materials: Aston,Bournemouth & Poole College, Cambridge, City, Glamorgan, Glasgow, Glasgow Caledonian, Glenrothes In-stitute of Applied Technology, Harper Adams University College, Hertfordshire, Leicester, Liverpool, LondonMetropolitan, Moray College, Northumbria, Nottingham, Nottingham Trent, Oxford Brookes, Plymouth,Portsmouth, Queens Belfast, Robert Gordon, Royal Forest of Dean College, Salford, Sligo Institute of Tech-nology, Southampton, Southampton Institute, Surrey, Teesside, Ulster, University of Wales Institute Cardiff,West Kingsway College (London), West Notts College.

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1 Basic Algebra 26 Functions of a Complex Variable2 Basic Functions 27 Multiple Integration3 Equations, Inequalities & Partial Fractions 28 Differential Vector Calculus4 Trigonometry 29 Integral Vector Calculus5 Functions and Modelling 30 Introduction to Numerical Methods6 Exponential and Logarithmic Functions 31 Numerical Methods of Approximation7 Matrices 32 Numerical Initial Value Problems8 Matrix Solution of Equations 33 Numerical Boundary Value Problems9 Vectors 34 Modelling Motion10 Complex Numbers 35 Sets and Probability11 Differentiation 36 Descriptive Statistics12 Applications of Differentiation 37 Discrete Probability Distributions13 Integration 38 Continuous Probability Distributions14 Applications of Integration 1 39 The Normal Distribution15 Applications of Integration 2 40 Sampling Distributions and Estimation16 Sequences and Series 41 Hypothesis Testing17 Conics and Polar Coordinates 42 Goodness of Fit and Contingency Tables18 Functions of Several Variables 43 Regression and Correlation19 Differential Equations 44 Analysis of Variance20 Laplace Transforms 45 Non-parametric Statistics21 z-Transforms 46 Reliability and Quality Control22 Eigenvalues and Eigenvectors 47 Mathematics and Physics Miscellany23 Fourier Series 48 Engineering Case Studies24 Fourier Transforms 49 Student’s Guide25 Partial Differential Equations 50 Tutor’s Guide

Copyright Loughborough University, 2006

ContentsContents !"!"Integral Vector

Calculus 29.1 Line Integrals 2

29.2 Surface and Volume Integrals 34

29.3 Integral Vector Theorems 55

LearningIn this Workbook you will learn how to integrate functions involving vectors. You will learnhow to evaluate line integrals i.e. where a scalar or a vector is summed along a line orcontour. You will be able to evaluate surface and volume integrals where a functioninvolving vectors is summed over a surface or volume. You will learn about some theoremsrelating to line, surface or volume integrals viz Stokes' theorem, Gauss' divergence theorem and Green's theorem.

outcomes

Surface and

Volume Integrals

29.2

Introduction

A vector or scalar field - including one formed from a vector derivative (div, grad or curl) - can beintegrated over a surface or volume. This Section shows how to carry out such operations.

Prerequisites

Before starting this Section you should . . .

• be familiar with vector derivatives

• be familiar with double and triple integrals

Learning Outcomes

On completion you should be able to . . .

• carry out operations involving integration ofscalar and vector fields

34 HELM (2008):

Workbook 29: Integral Vector Calculus

®

1. Surface integrals involving vectors

The unit normal

For the surface of any three-dimensional shape, it is possible to find a vector lying perpendicular tothe surface and with magnitude 1. The unit vector points outwards from a closed surface and isusually denoted by n.

Example 17

If S is the surface of the sphere x2 + y2 + z2 = a2 find the unit normal n.

Solution

The unit normal at the point P (x, y, z) points away from the centre of the sphere i.e. it lies inthe direction of xi + yj + zk. To make this a unit vector it must be divided by its magnitude

x2 + y2 + z2 i.e. the unit vector is

n =x

x2 + y2 + z2i +

yx2 + y2 + z2

j +z

x2 + y2 + z2k

=x

ai +

y

aj +

z

ak

where a =

x2 + y2 + z2 is the radius of the sphere.

z

x

y

n

i

j

k

P (x, y, z)

a

Figure 6: A unit normal n to a sphere

HELM (2008):

Section 29.2: Surface and Volume Integrals

35

®

Integral Vector

Theorems

29.3

Introduction

Various theorems exist relating integrals involving vectors. Those involving line, surface and volumeintegrals are introduced here.

They are the multivariable calculus equivalent of the fundamental theorem of calculus for singlevariables (“integration and differentiation are the reverse of each other”).

Use of these theorems can often make evaluation of certain vector integrals easier. This Sectionintroduces the main theorems which are Gauss’ divergence theorem, Stokes’ theorem and Green’stheorem.

Prerequisites


• be able to find the gradient of a scalar fieldand the divergence and curl of a vector field

• be familiar with the integration of vectorfunctions

Learning Outcomes


• use vector integral theorems to facilitatevector integration

HELM (2008):

Section 29.3: Integral Vector Theorems

55

1. Stokes’ theorem

This is a theorem that equates a line integral to a surface integral. For any vector field F and acontour C which bounds an area S,

S

(∇× F ) · dS =

C

F · dr

S

S

C

d

Figure 16: A surface for Stokes’ theoremNotes

(a) dS is a vector perpendicular to the surface S and dr is a line element along the contour C.The sense of dS is linked to the direction of travel along C by a right hand screw rule.

(b) Both sides of the equation are scalars.

(c) The theorem is often a useful way of calculating a line integral along a contour composed ofseveral distinct parts (e.g. a square or other figure).

(d) ∇ × F is a vector field representing the curl of the vector field F and may, alternatively, bewritten as curl F .

Justification of Stokes’ theorem

Imagine that the surface S is divided into a set of infinitesimally small rectangles ABCD where theaxes are adjusted so that AB and CD lie parallel to the new x-axis i.e. AB = δx and BC and ADlie parallel to the new y-axis i.e. BC = δy.

Now,

C

F · dr is calculated, where C is the boundary of a typical such rectangle.

The contributions along AB, BC, CD and DA are

F (x, y, 0) · δx = Fx(x, y, z)δx,

F (x + δx, y, 0) · δy = Fy(x + δx, y, z)δy,

F (x, y + δy, 0) · (−δx) = −Fx(x, y + δy, z)δx

F (x, y, 0) · (−δx) = −Fy(x, y, z)δy.

Thus,

C

F · dr ≈ (Fx(x, y, z)− Fx(x, y + δy, z))δx + (Fy(x + δx, y, z)− Fy(x, y, z))δy

≈ ∂Fy

∂xδxδy − ∂Fx

∂yδxδy

≈ (∇× F )zδS

= (∇× F ) · dS

as dS is perpendicular to the x- and y- axes.

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Thus, for each small rectangle,

C

F · dr ≈ (∇× F ) · dS

When the contributions over all the small rectangles are summed, the line integrals along the innerparts of the rectangles cancel and all that remains is the line integral around the outside of the surfaceS. The surface integrals sum. Hence, the theorem applies for the area S bounded by the contour C.While the above does not constitute a formal proof of Stokes’ theorem, it does give an appreciationof the origin of the theorem.

Contribution doesnot cancel

Contributions cancel

Figure 17: Line integral cancellation and non-cancellation

Key Point 8

Stokes’ Theorem

C

F · dr =

S

(∇× F ) · dS

The closed contour integral of the scalar product of a vector function with the vector along thecontour is equal to the integral of the scalar product of the curl of that vector function and the unitnormal, over the corresponding surface.

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Example 32

Verify Stokes’ theorem for the vector function F = y2i− (x + z)j + yzk and theunit square 0 ≤ x ≤ 1, 0 ≤ y ≤ 1, z = 0.

Solution

If F = y2i− (x + z)j + yzk then ∇× F = (z + 1)i + (−1− 2y)k = i + (−1− 2y)k (as z = 0).Note that dS = dxdyk so that (∇× F ) · dS = (−1− 2y)dydx

Thus

S

(∇× F ) · dS =

1

x=0

1

y=0

(−1− 2y)dydx

=

1

x=0

(−y − y2)

1

y=0

dx =

1

x=0

(−2)dx

=

− 2x

1

0

= −2 + 0 = −2

To evaluate

C

F · dr, we must consider the four sides separately.

When y = 0, F = −xj and dr = dxi so F · dr = 0 i.e. the contribution of this side to the integralis zero.When x = 1, F = y2i− j and dr = dyj so F · dr = −dy so the contribution to the integral is

1

y=0

(−dy) =

− y

1

0

= −1.

When y = 1, F = i− xj and dr = −dxi so F · dr = −dx so the contribution to the integral is 1

x=0

(−dx) =

− x

1

0

= −1.

When x = 0, F = y2i and dr = −dyj so F · dr = 0 so the contribution to the integral is zero.

The integral

C

F · dr is the sum of the contributions i.e. 0− 1− 1 + 0 = −2.

Thus

S

(∇× F ) · dS =

C

F · dr = −2 i.e. Stokes’ theorem has been verified.

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Example 33

Using cylindrical polar coordinates verify Stokes’ theorem for the function F = ρ2φthe circle ρ = a, z = 0 and the surface ρ ≤ a, z = 0.

Solution

Firstly, find

C

F · dr. This can be done by integrating along the contour ρ = a from φ = 0 to

φ = 2π. Here F = a2φ (as ρ = a) and dr = a dφ φ (remembering the scale factor) so F ·dr = a3dφand hence

C

F · dr =

2π

0

a3dφ = 2πa3

As F = ρ2φ, ∇× F = 3ρz and (∇× F ) · dS = 3ρ as dS = z.Thus

S

(∇× F ) · dS =

2π

φ=0

1

ρ=0

3ρ× ρdρdφ =

2π

φ=0

a

ρ=0

3ρ2dρdφ

=

2π

φ=0

ρ3

a

ρ=0

dφ =

2π

0

a3dφ = 2πa3

Hence

C

F · dr =

S

(∇× F ) · dS = 2πa3

Example 34

Find the closed line integral

C

F ·dr for the vector field F = y2i+(x2−z)j+2xyk

and for the contour ABCDEFGHA in Figure 18.

A(0, 0)

H(0, 4)

F (1, 7)

G(1, 4)

E(5, 7)

D(2, 4)C(6, 4)

B(6, 0) x

y

Figure 18: Closed contour ABCDEFGHA

HELM (2008):


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Solution

To find the line integral directly would require eight line integrals i.e. along AB, BC, CD, DE,

EF , FG, GH and HA. It is easier to carry out a surface integral to find

S

(∇× F ) · dS which

is equal to the required line integral

C

F · dr by Stokes’ theorem.

As F = y2i + (x2 − z)j + 2xyk, ∇× F =

i j k∂∂x

∂∂y

∂∂z

y2 x2 − z 2xy

= (2x + 1)i− 2yj + (2x− 2y)k

As the contour lies in the x-y plane, the unit normal is k and dS = dxdykHence (∇× F ) · dS = (2x− 2y)dxdy.

To work out

S

(∇×F ) · dS, it is necessary to divide the area inside the contour into two smaller

areas i.e. the rectangle ABCDGH and the trapezium DEFG. On ABCDGH, the integral is

4

y=0

6

x=0

(2x− 2y)dxdy =

4

y=0

x2 − 2xy

6

x=0dy =

4

y=0

(36− 12y)dy

=

36y − 6y2

4

0

= 36× 4− 6× 16− 0 = 48

On DEFG, the integral is

7

y=4

y−2

x=1

(2x− 2y)dxdy =

7

y=4

x2 − 2xy

y−2

x=1

dy =

7

y=4

(−y2 + 2y + 3)dy

=

−1

3y3 + y2 + 3y

7

4

= −343

3+ 49 + 21 +

64

3− 16− 12 = −51

So the full integral is,

S

(∇× F ) · dS = 48− 51 = −3.

∴ By Stokes’ theorem,

C

F · dr = −3

From Stokes’ theorem, it can be seen that surface integrals of the form

S

(∇× F ) · dS depend

only on the contour bounding the surface and not on the internal part of the surface.

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Task

Verify Stokes’ theorem for the vector field F = x2i + 2xyj + zk and the trianglewith vertices at (0, 0, 0), (3, 0, 0) and (3, 1, 0).

First find the normal vector dS:

Your solution

Answer

dxdyk

Then find the vector ∇× F :

Your solution

Answer

2yk

Now evaluate the double integral

S

(∇× F ) · dS over the triangle:

Your solution

Answer

1

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61

Finally find the integral

F · dr along the 3 sides of the triangle and so verify that the two sides of

the Stokes’ theorem are equal:

Your solution

Answer

9 + 3− 11 = 1, Both sides of Stokes’ theorem have value 1.

Exercises

1. Using plane-polar coordinates (or cylindrical polar coordinates with z = 0), verify Stokes’

theorem for the vector field F = ρρ + ρ cosπρ

2

φ and the semi-circle ρ ≤ 1, −π

2 ≤ φ ≤ π2 .

2. Verify Stokes’ theorem for the vector field F = 2xi + (y2− z)j + xzk and the contour aroundthe rectangle with vertices at (0,−2, 0),(2,−2, 0), (2, 0, 1) and (0, 0, 1).

3. Verify Stokes’ theorem for the vector field F = −yi + xj + zk

(a) Over the triangle (0, 0, 0), (1, 0, 0), (1, 1, 0).

(b) Over the triangle (1, 0, 0), (1, 1, 0), (1, 1, 1).

4. Use Stokes’ theorem to evaluate the integral

C

F · dr where F =

sin(

1

x+ 1) + 5y

i + (2x− ey2

)j

and C is the contour starting at (0, 0) and going to (5, 0), (5, 2), (6, 2), (6, 5), (3, 5), (3, 2),(0, 2) and returning to (0, 0).

Answers

1. Both integrals give 0,

2. Both integrals give 1

3. (a) Both integrals give 1 (b) Both integrals give 0 (as ∇× F is perpendicular to dS)

4. −57, [∇× F = −3 k].

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2. Gauss’ theorem

This is sometimes known as the divergence theorem and is similar in form to Stokes’ theorem butequates a surface integral to a volume integral. Gauss’ theorem states that for a volume V , boundedby a closed surface S, any ‘well-behaved’ vector field F satisfies

S

F · dS =

V

∇ · F dV

Notes:

(a) dS is a unit normal pointing outwards from the interior of the volume V .

(b) Both sides of the equation are scalars.

(c) The theorem is often a useful way of calculating a surface integral over a surface composed ofseveral distinct parts (e.g. a cube).

(d) ∇ · F is a scalar field representing the divergence of the vector field F and may, alternatively,be written as div F .

(e) Gauss’ theorem can be justified in a manner similar to that used for Stokes’ theorem (i.e. byproving it for a small volume element, then summing up the volume elements and allowing theinternal surface contributions to cancel.)

Key Point 9Gauss’ Theorem

S

F · dS =

V

∇ · FdV

The closed surface integral of the scalar product of a vector function with the unit normal (or flux ofa vector function through a surface) is equal to the integral of the divergence of that vector functionover the corresponding volume.

HELM (2008):


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Example 35

Verify Gauss’ theorem for the unit cube 0 ≤ x ≤ 1, 0 ≤ y ≤ 1, 0 ≤ z ≤ 1 andthe function F = xi + zj

Solution

To find

S

F · dS, the integral must be evaluated for all six faces of the cube and the results

summed.On the face x = 0, F = zj and dS = −i dydz so F · dS = 0 and

S

F · dS =

1

0

1

0

0 dydz = 0

On the face x = 1, F = i + zj and dS = i dydz so F · dS = 1 dydz and

S

F · dS =

1

0

1

0

1 dydz = 1

On the face y = 0, F = xi + zj and dS = −j dxdz so F · dS = −z dxdz and

S

F · dS = − 1

0

1

0

z dxdz = −1

2

On the face y = 1, F = xi + zj and dS = j dxdz so F · dS = z dxdz and

S

F · dS =

1

0

1

0

z dxdz =1

2

On the face z = 0, F = xi and dS = −k dydz so F · dS = 0 dxdy and

S

F · dS =

1

0

1

0

0 dxdy = 0

On the face z = 1, F = xi + j and dS = k dydz so F · dS = 0 dxdy and

S

F · dS =

1

0

1

0

0 dxdy = 0

Thus, summing over all six faces,

S

F · dS = 0 + 1− 1

2+

1

2+ 0 + 0 = 1.

To find

V

∇ · F dV note that ∇ · F =∂

∂xx +

∂

∂yz = 1 + 0 = 1.

So

V

∇ · F dV =

1

0

1

0

1

0

1 dxdydz = 1.

So

S

F · dS =

V

∇ · F dV = 1 hence verifying Gauss’ theorem.

Note: The volume integral needed just one triple integral, but the surface integral required six doubleintegrals. Reducing the number of integrals is often the motivation for using Gauss’ theorem.

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Example 36

Use Gauss’ theorem to evaluate the surface integral

S

F · dS where F is the

vector field x2yi + 2xyj + z3k and S is the surface of the unit cube 0 ≤ x ≤ 1,0 ≤ y ≤ 1, 0 ≤ z ≤ 1.

Solution

Note that to carry out the surface integral directly will involve, as in Example 35, the evaluation ofsix double integrals. However, by Gauss’ theorem, the same result comes from the volume integral

V

∇ · F dV . As ∇ · F = 2xy + 2x + 3z2, we have the triple integral

1

0

1

0

1

0

(2xy + 2x + 3z2) dxdydz

=

1

0

1

0

x2y + x2 + 3xz2

1

x=0

dydz =

1

0

1

0

(y + 1 + 3z2)dydz

=

1

0

1

2y2 + y + 3yz2

1

y=0

dz =

1

0

(1

2+ 1 + 3z2)dz =

1

0

(3

2+ 3z2)dz

=

3

2z + z3

1

0

=5

2

The six double integrals would also sum to 52 but this approach would require much more effort.

Engineering Example 5

Gauss’ law

Introduction

From Gauss’ theorem, it is possible to derive a result which can be used to gain insight into situationsarising in Electrical Engineering. Knowing the electric field on a closed surface, it is possible to findthe electric charge within this surface. Alternatively, in a sufficiently symmetrical situation, it ispossible to find the electric field produced by a given charge distribution.Gauss’ theorem states

S

F · dS =

V

∇ · F dV

If F = E, the electric field, it can be shown that,

∇ · F = ∇ · E =q

ε0

HELM (2008):


65

where q is the amount of charge per unit volume, or charge density, and ε0 is the permittivity of freespace: ε0 = 10−9/36π F m−1 ≈ 8.84×10−12 F m−1. Gauss’ theorem becomes in this case

S

E · dS =

V

∇ · E dV =

V

q

ε0dV =

1

ε0

V

q dV =Q

ε0

i.e.

S

E · dS =Q

ε0

which is known as Gauss’ law. Here Q is the total charge inside the surface S.

Note: this is one of the important Maxwell’s Laws.

Problem in words

A point charge lies at the centre of a cube. Given the electric field, find the magnitude of the charge,using Gauss’ law .

Mathematical statement of problem

Consider the cube −12 ≤ x ≤ 1

2 , −12 ≤ y ≤ 1

2 , −12 ≤ z ≤ 1

2 where the dimensions are in metres. Apoint charge Q lies at the centre of the cube. If the electric field on the top face (z = 1

2) is given by

E = 10xi + yj + zk

(x2 + y2 + z2)32

find the charge Q from Gauss’ law .

Hint :

12

x=− 12

12

y=− 12

x2 + y2 +

1

4

− 32

dy dx =4π

3

Mathematical analysis

From Gauss’ law

S

E · dS =Q

ε0

so

Q = ε0

S

E · dS = 6ε0

S(top)

E · dS

since, using the symmetry of the six faces of the cube, it is possible to integrate over just one ofthem (here the top face is chosen) and multiply by 6. On the top face

E = 10xi + yj + 1

2kx2 + y2 + 1

4

32

and

dS = (element of surface area)× (unit normal)

= dx dy k

66 HELM (2008):


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So

E · dS = 1012

x2 + y2 + 14

32

dy dx

= 5

x2 + y2 +

1

4

− 32

dy dx

Now

S(top)

E · dS =

12

x=− 12

12

y=− 12

5

x2 + y2 +

1

4

− 32

dy dx

= 5× 4π

3(using the hint)

=20π

3

So, from Gauss’ law,

Q = 6ε0 ×20π

3= 40πε0 ≈ 10−9C

Interpretation

Gauss’ law can be used to find a charge from its effects elsewhere.

The form of E = 10xi + yj + 1

2kx2 + y2 + 1

4

32

comes from the fact that E is radial and equals 10r

r3= 10

r

r2

Example 37

Verify Gauss’ theorem for the vector field F = y2j− xzk and the triangular prismwith vertices at (0, 0, 0), (2, 0, 0), (0, 0, 1), (0, 4, 0), (2, 4, 0) and (0, 4, 1) (seeFigure 19).

(0, 0, 0)

(0, 0, 1)

(2, 0, 0)

(0, 4, 0)(2, 4, 0)

(0, 4, 1)

x

yz

Figure 19: The triangular prism defined by six vertices

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Solution

As F = y2j − xzk, ∇ · F = 0 + 2y − x = 2y − x.Thus

V

∇ · FdV =

2

x=0

4

y=0

1−x/2

z=0

(2y − x)dzdydx

=

2

x=0

4

y=0

2yz − xz

1−x/2

z=0

dydx =

2

x=0

4

y=0

(2y − xy − x +1

2x2)dydx

=

2

x=0

y2 − 1

2xy2 − xy +

1

2x2y

4

y=0

dx =

2

x=0

(16− 12x + 2x2)dx

=

16x− 6x2 +

2

3x3

2

0

=40

3

To work out

S

F · dS, it is necessary to consider the contributions from the five faces separately.

On the front face, y = 0, F = −xzk and dS = −j thus F · dS = 0 and the contribution to theintegral is zero.On the back face, y = 4, F = 16j − xzk and dS = j thus F · dS = 16 and the contribution to theintegral is

2

x=0

1−x/2

z=0

16dzdx =

2

x=0

16z

1−x/2

z=0

dx =

2

x=0

16(1− x/2)dx =

16x− 4x2

2

0

= 16.

On the left face, x = 0, F = y2j and dS = −i thus F · dS = 0 and the contribution to the integralis zero.On the bottom face, z = 0, F = y2j and dS = −k thus F · dS = 0 and the contribution to theintegral is zero.On the top right (sloping) face, z = 1−x/2, F = y2j+(1

2x2−x)k and the unit normal n = 1√

5i+ 2√

5k

Thus dS =

1√5i + 2√

5kdydw where dw measures the distance along the slope for a constant y.

As dw =√

52 dx, dS =

12i + k

dydx thus F · dS = 16 and the contribution to the integral is

2

x=0

4

y=0

(1

2x2 − x)dydx =

2

x=0

(2x2 − 4x)dx =

2

3x3 − 2x2

2

0

= −8

3.

Adding the contributions,

S

F · dS = 0 + 16 + 0 + 0− 8

3=

40

3.

Thus

S

F · dS =

V

∇ · FdV =40

3hence verifying Gauss’ divergence theorem.

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Field strength around a charged line

Problem in words

Find the electric field strength at a given distance from a uniformly charged line.


Determine the electric field at a distance r from a uniformly charged line (charge per unit length ρL).You may assume from symmetry that the field points directly away from the line.

l

r

Figure 20: Field strength around a line charge


Imagine a cylinder a distance r from the line and of length l (see Figure 20). From Gauss’ law

S

E · dS =Q

ε0

As the charge per unit length is ρL, then the right-hand side equals ρLl/ε0. On the left-hand side,the integral can be expressed as the sum

S

E · dS =

S(ends)

E · dS +

S(curved)

E · dS

Looking first at the circular ends of the cylinder, the fact that the field lines point radially awayfrom the charged line implies that the electric field is in the plane of these circles and has no normalcomponent. Therefore E · dS will be zero for these ends.Next, over the curved surface of the cylinder, the electric field is normal to it, and the symmetryof the problem implies that the strength of the electric field will be constant (here denoted by E).Therefore the integral = Total curved surface area × Field strength = 2πrlE.

So, by Gauss’ law

S(ends)

E · dS +

S(curved)

E · dS =Q

ε0

or

0 + 2πrlE =ρLl

ε0

Interpretation

Hence, the field strength E is given by E =ρL

2πε0r

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Field strength on a cylinder

Problem in words

Given the electric field E on the surface of a cylinder, use Gauss’ law to find the charge per unitlength.


On the surface of a long cylinder of radius a and length l, the electric field is given by

E =ρL

2πε0

(a + b cos θ) r − b sin θ θ

(a2 + 2ab cos θ + b2)

(using cylindrical polar co-ordinates) due to a line of charge a distance b (< a) from the centre ofthe cylinder. Using Gauss’ law , find the charge per unit length.

Hint:-

2π

0

a + b cos θ

(a2 + 2ab cos θ + b2)dθ =

2π

a


Consider a cylindrical section - as in the previous example, there are no contributions from the endsof the cylinder since the electric field has no normal component here. However, on the curved surface

dS = a dθ dz r

so

E · dS =ρL

2πε0

a + b cos θ

(a2 + 2ab cos θ + b2)a dθ dz

Integrating over the curved surface of the cylinder

S

E · dS =

l

z=0

θ=2π

θ=0

aρL

2πε0

a + b cos θ

(a2 + 2ab cos θ + b2)dθ dz

=aρLl

2πε0

2π

0

a + b cos θ

(a2 + 2ab cos θ + b2)dθ

=ρLl

ε0using the given result for the integral.

Then, if Q is the total charge inside the cylinder, from Gauss’ law

ρLl

ε0=

Q

ε0so ρL =

Q

las one would expect.

Interpretation

Therefore the charge per unit length on the line of charge is given by ρL (i.e. the charge per unitlength is constant).

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Task

Verify Gauss’ theorem for the vector field F = xi − yj + zk and the unit cube0 ≤ x ≤ 1, 0 ≤ y ≤ 1, 0 ≤ z ≤ 1.

(a) Find the vector ∇ · F .

(b) Evaluate the integral

1

z=0

1

y=0

1

x=0

∇ · Fdxdydz.

(c) For each side, evaluate the normal vector dS and the surface integral

S

F · dS.

(d) Show that the two sides of the statement of Gauss’ theorem are equal.

Your solution

Answer(a) 1− 1 + 1 = 1

(b) 1

(c) −dxdyk, 0; dxdyk, 1; −dxdzj, 0; dxdzj, −1; −dydzi, 0; dydzi, 1

(d) Both sides are equal to 1.

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Exercises

1. Verify Gauss’ theorem for the vector field F = 4xzi − y2j + yzk and the cuboid 0 ≤ x ≤ 2,0 ≤ y ≤ 3, 0 ≤ z ≤ 4.

2. Verify Gauss’ theorem, using cylindrical polar coordinates, for the vector field F = ρ−2ρ overthe cylinder 0 ≤ ρ ≤ r0, −1 ≤ z ≤ 1 for

(a) r0 = 1

(b) r0 = 2

3. If S is the surface of the tetrahedron with vertices at (0, 0, 0), (1, 0, 0), (0, 1, 0) and (0, 0, 1),find the surface integral

S

(xi + yzj) · dS

(a) directly

(b) by using Gauss’ theorem

Hint :- When evaluating directly, show that the unit normal on the sloping face is 1√3(i+ j +k)

and that dS = (i + j + k)dxdy

Answers

1. Both sides are 156,

2. Both sides equal (a) 4π, (b) 2π,

3. (a)5

24[only contribution is from the sloping face] (b)

5

24[by volume integral of (1 + z)].

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3. Green’s Identities (3D)

Like Gauss’ theorem, Green’s identities relate surface integrals to volume integrals. However, Green’sidentities are concerned with two scalar fields u(x, y, z) and w(x, y, z). Two statements of Green’sidentities are as follows

S

(u∇w) · dS =

V

∇u ·∇w + u∇2w

dV [1]

and

S

u∇w − v∇u · dS =

V

u∇2w − w∇2u

dV [2]

Proof of Green’s identities

Green’s identities can be derived from Gauss’ theorem and a vector derivative identity.

Vector identity (1) from subsection 6 of 28.2 states that ∇ · (φA) = (∇φ) · A + φ(∇ · A).

Letting φ = u and A = ∇w in this identity,

∇ · (u∇w) = (∇u) · (∇w) + u(∇ · (∇w)) = (∇u) · (∇w) + u∇2w

Gauss’ theorem states

S

F · dS =

V

∇ · FdV

Now, letting F = u∇w,

S

(u∇w) · dS =

V

∇ · (u∇w)dV

=

V

(∇u) · (∇w) + u∇2w

dV

This is Green’s identity [1].

Reversing the roles of u and w,

S

(w∇u) · dS =

V

(∇w) · (∇u) + w∇2u

dV

Subtracting the last two equations yields Green’s identity [2].

Key Point 10

Green’s Identities

[1]

S

(u∇w) · dS =

V

∇u ·∇w + u∇2w

dV

[2]

S

u∇w − v∇u · dS =

V

u∇2w − w∇2u

dV

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Example 38

Verify Green’s first identity for u = (x − x2)y, w = xy + z2 and the unit cube,0 ≤ x ≤ 1, 0 ≤ y ≤ 1, 0 ≤ z ≤ 1.

Solution

As w = xy + z2, ∇w = yi + xj + 2zk. Thus u∇w = (xy − x2y)(yi + xj + 2zk) and the surfaceintegral is of this quantity (scalar product with dS) integrated over the surface of the unit cube.

On the three faces x = 0, x = 1, y = 0, the vector u∇w = 0 and so the contribution to the surfaceintegral is zero.

On the face y = 1, u∇w = (x−x2)(i+xj +2zk) and dS = dxdzj so (u∇w) ·dS = (x2−x3)dxdzand the contribution to the integral is

1

x=0

1

z=0

(x2 − x3)dzdx =

1

0

(x2 − x3)dx =

x3

3− x4

4

1

0

=1

12.

On the face z = 0, u∇w = (x − x2)y(yi + xj) and dS = −dxdzk so (u∇w) · dS = 0 and thecontribution to the integral is zero.

On the face z = 1, u∇w = (x−x2)y(yi+xj+2k) and dS = dxdyk so (u∇w)·dS = 2y(x−x2)dxdyand the contribution to the integral is 1

x=0

1

y=0

2y(x− x2)dydx =

1

x=0

y2(x− x2)

1

y=0

dx =

1

0

(x− x2)dx =1

6.

Thus,

S

(u∇w) · dS = 0 + 0 + 0 +1

12+ 0 +

1

6=

1

4.

Now evaluate

V

∇u ·∇w + u∇2w

dV .

Note that ∇u = (1− 2x)yi + (x− x2)j and ∇2w = 2 so

∇u ·∇w + u∇2w = (1− 2x)y2 + (x− x2)x + 2(x− x2)y = x2 − x3 + 2xy − 2x2y + y2 − 2xy2

and the integral

V

∇u ·∇w + u∇2w

dV =

1

z=0

1

y=0

1

x=0

(x2 − x3 + 2xy − 2x2y + y2 − 2xy2)dxdydz

=

1

z=0

1

y=0

x3

3− x4

4+ x2y − 2

3x3y + xy2 − x2y2

1

x=0

dydz

=

1

z=0

1

y=0

(1

12+

y

3)dydz =

1

z=0

y

12+

y2

6

1

y=0

dz

=

1

z=0

(1

4)dz =

z

4

1

z=0=

1

4

Hence

S

(u∇w) ·dS =

V

∇u ·∇w + u∇2w

dV =

1

4and Green’s first identity is verified.

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Green’s theorem in the plane

This states that

C

(Pdx + Qdy) =

S

∂Q

∂x− ∂P

∂y

dxdy

S is a 2-D surface with perimeter C; P (x, y) and Q(x, y) are scalar functions.

This should not be confused with Green’s identities.

Justification of Green’s theorem in the plane

Green’s theorem in the plane can be derived from Stokes’ theorem.

S

(∇× F ) · dS =

C

F · dr

Now let F be the vector field P (x, y)i + Q(x, y)j i.e. there is no dependence on z and there are nocomponents in the z− direction. Now

∇× F =

i j k

∂

∂x

∂

∂y

∂

∂z

P (x, y) Q(x, y) 0

=

∂Q

∂x− ∂P

∂y

k

and dS = dxdyk giving (∇× F ) · dS =

∂Q

∂x− ∂P

∂y

dxdy.

Thus Stokes’ theorem becomes

S

∂Q

∂x− ∂P

∂y

dxdy =

C

F · dr

and Green’s theorem in the plane follows.

Key Point 11Green’s Theorem in the Plane

C

(Pdx + Qdy) =

S

∂Q

∂x− ∂P

∂y

dxdy

This relates a line integral around a closed path C with a double integral over the region S enclosedby C. It is effectively a two-dimensional form of Stokes’ theorem.

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Example 39

Evaluate the line integral

C

(4x2 + y − 3)dx + (3x2 + 4y2 − 2)dy

around the

rectangle 0 ≤ x ≤ 3, 0 ≤ y ≤ 1.

Solution

The integral could be obtained by evaluating four line integrals but it is easier to note that[(4x2 + y − 3)dx + (3x2 + 4y2 − 2)dy] is of the form Pdx + Qdy with P = 4x2 + y − 3 andQ = 3x2 + 4y2 − 2. It is thus of a suitable form for Green’s theorem in the plane.

Note that∂Q

∂x= 6x and

∂P

∂y= 1.

Green’s theorem in the plane becomes

C

(4x2 + y − 3)dx + (3x2 + 4y2 − 2)dy =

1

y=0

3

x=0

(6x− 1) dxdy

=

1

y=0

3x2 − x

3

x=0

dy =

1

y=0

24 dy = 24

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Example 40

Verify Green’s theorem in the plane for the integral

C

4zdy + (y2 − 2)dz

and

the triangular contour starting at the origin O = (0, 0, 0) and going to A = (0, 2, 0)and B = (0, 0, 1) before returning to the origin.

Solution

The whole of the contour is in the plane x = 0 and Green’s theorem in the plane becomes

C

(Pdy + Qdz) =

S

∂Q

∂y− ∂P

∂z

dydz

(a) Firstly evaluate

C

4zdy + (y2 − 2)dz

.

On OA, z = 0 and dz = 0. As the integrand is zero, the integral will also be zero.On AB, z = (1− y

2) and dz = −12dy. The integral is 0

y=2

(4− 2y)dy − 1

2(y2 − 2)dy

=

0

2

(5− 2y − 1

2y2)dy =

5y − y2 − 1

6y3

0

2

= −14

3

On BO, y = 0 and dy = 0. The integral is

0

1

(−2)dz =

− 2z

0

1

= 2.

Summing,

C

4zdy + (y2 − 2)dz

= −8

3

(b) Secondly evaluate

S

∂Q

∂y− ∂P

∂z

dydz

In this example, P = 4z and Q = y2 − 2. Thus∂P

∂z= 4 and

∂Q

∂y= 2y. Hence,

S

∂Q

∂y− ∂P

∂z

dydz =

2

y=0

1−y/2

z=0

(2y − 4) dzdy

=

2

y=0

2yz − 4z

1−y/2

z=0

dy =

2

y=0

−y2 + 4y − 4

dy

=

−1

3y3 + 2y2 − 4y

2

0

= −8

3

Hence:

C

(Pdy + Qdz) =

S

∂Q

∂y− ∂P

∂z

dydz = −8

3and Green’s theorem in the plane is verified.

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One very useful, special case of Green’s theorem in the plane is when Q = x and P = −y. Thetheorem becomes

C

−ydx + xdy =

S

(1− (−1)) dxdy

The right-hand side becomes

S

2 dxdy i.e. 2A where A is the area inside the contour C. Hence

A =1

2

C

xdy − ydx

This result is known as the area theorem. It gives us the area bounded by a curve C in terms of aline integral around C.

Example 41

Verify the area theorem for the segment of the circle x2 + y2 = 4 lying above theline y = 1.

Solution

Firstly, the area of the segment ADBC can be found by subtracting the area of the triangle OADBfrom the area of the sector OACB. The triangle has area 1

2 × 2√

3× 1 =√

3. The sector has areaπ3 × 22 = 4

3π. Thus segment ADBC has area 43π −

√3.

Now, evaluate the integral

C

xdy − ydx around the segment.

Along the line, y = 1, dy = 0 so the integral

C

xdy − ydx becomes

√3

−√

3

(x × 0 − 1 × dx) =

√3

−√

3

(−dx) = −2√

3.

Along the arc of the circle, y =√

4− x2 = (4 − x2)1/2 so dy = −x(4 − x2)−1/2dx. The integral

C

xdy − ydx becomes

−√

3

√3

−x2(4− x2)−1/2 − (4− x2)1/2dx =

√3

−√

3

4√4− x2

dx

=

π/3

−π/3

41

2 cos θ2 cos θ dθ (letting x = 2 sin θ)

=

π/3

−π/3

4dθ =8

3π

So, 12

C

xdy − ydx =1

2

8

3π − 2

√3

=

4

3π −

√3.

Hence both sides of the area theorem equal 43π −

√3 thus verifying the theorem.

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Task

Verify Green’s theorem in the plane when applied to the integral

C

(5x + 2y − 7)dx + (3x− 4y + 5)dy

where C represents the perimeter of the trapezium with vertices at (0, 0), (3, 0),(6, 1) and (1, 1).

First let P = 5x + 2y − 7 and Q = 3x− 4y + 5 and find∂Q

∂x− ∂P

∂y:

Your solution

Answer

1

Now find

∂Q

∂x− ∂P

∂y

dxdy over the trapezium:

Your solution

Answer

4 (by elementary geometry)

Now find

(Pdx + Qdy) along the four sides of the trapezium, beginning with the line from (0, 0)

to (3, 0), and then proceeding anti-clockwise.

Your solution

Answers 1.5, 66, −62.5, −1 whose sum is 4.

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79

Finally show that the two sides of the statement of Green’s theorem are equal:

Your solution

Answer

Both sides are 4.

Exercises

1. Verify Green’s identity [1] (page 73) for the functions u = xyz, w = y2 and the unit cube0 ≤ x ≤ 1, 0 ≤ y ≤ 1, 0 ≤ z ≤ 1.

2. Verify the area theorem for

(a) The area above y = 0, but below y = 1− x2.

(b) The segment of the circle x2 + y2 = 1, to the upper left of the line y = 1− x.

Answers

1. Both integrals in [1] equal1

2

2. (a) both sides give a value of4

3, (b) both sides give a value of

π

4− 1

2.

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Example 18

For the cube 0 ≤ x ≤ 1, 0 ≤ y ≤ 1, 0 ≤ z ≤ 1, find the unit outward normal n

for each face.

Solution

On the face given by x = 0, the unit normal points in the negative x-direction. Hence the unitnormal is −i. Similarly :-On the face x = 1 the unit normal is i. On the face y = 0 the unit normal is −j.On the face y = 1 the unit normal is j. On the face z = 0 the unit normal is −k.On the face z = 1 the unit normal is k.

dddSSS and the unit normal

The vector dS is a vector, being an element of the surface with magnitude du dv and directionperpendicular to the surface.

If the plane in question is the Oxy plane, then dS = n du dv = k dx dy.

du dv

dS

uv

Figure 7: The vector dS as an element of a surface, with magnitude du dv

If the plane in question is not one of the three coordinate planes (Oxy, Oxz, Oyz), appropriateadjustments must be made to express dS in terms of two of dx and dy and dz.

Example 19

The rectangle OABC lies in the plane z = y (Figure 8).The vertices are O = (0, 0, 0), A = (1, 0, 0), B = (1, 1, 1) and C = (0, 1, 1).Find a unit vector n normal to the plane and an appropriate vector dS expressedin terms of dx and dy.

z

x

y

OA

BC

DE

Figure 8: The plane z = y passing through OABC

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Solution

Note that two vectors in the rectangle are−→OA = i and

−→OC = j + k. A vector perpendicular to the

plane is i× (j + k) = −j + k. However, this vector is of magnitude√

2 so the unit normal vector

is n =1√2(−j + k) = − 1√

2j +

1√2k.

The vector dS is therefore (− 1√2j+

1√2k) du dv where du and dv are increments in the plane of the

rectangle OABC. Now, one increment, say du, may point in the x-direction while dv will point in adirection up the plane, parallel to OC. Thus du = dx and (by Pythagoras) dv =

(dy)2 + (dz)2.

However, as z = y, dz = dy and hence dv =√

2dy.

Thus, dS = (− 1√2j +

1√2k) dx

√2 dy = (−j + k) dx dy.

Note :- the factor of√

2 could also have been found by comparing the area of rectangle OABC,

i.e. 1, with the area of its projection in the Oxy plane i.e. OADE with area1√2.

Integrating a scalar field

A function can be integrated over a surface by constructing a double integral and integrating in amanner similar to that shown in 27.1 and 27.2. Often, such integrals can be carried outwith respect to an element containing the unit normal.

Example 20

Evaluate the integral

A

1

1 + x2dS

over the area A where A is the square 0 ≤ x ≤ 1, 0 ≤ y ≤ 1, z = 0.

Solution

In this integral, dS becomes k dx dy i.e. the unit normal times the surface element. Thus theintegral is

1

y=0

1

x=0

k

1 + x2dx dy = k

1

y=0

tan−1

x

1

0dy

= k

1

y=0

(π

4− 0)

1

0dy =

π

4k

1

y=0

dy

=π

4k

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Example 21

Find

S

u dS where u = x2 + y2 + z2 and S is the surface of the unit cube

0 ≤ x ≤ 1, 0 ≤ y ≤ 1, 0 ≤ z ≤ 1.

Solution

The unit cube has six faces and the unit normal vector n points in a different direction on each face;see Example 18. The surface integral must be evaluated for each face separately and the resultssummed.On the face x = 0, the unit normal n = −i and the surface integral is

1

y=0

1

z=0

(02 + y2 + z

2)(−i)dzdy = −i

1

y=0

y

2z +

1

3z

3

1

z=0

dy

= −i

1

y=0

y

2 +1

3

dy = −i

1

3y

3 +1

3y

1

0

= −2

3i

On the face x = 1, the unit normal n = i and the surface integral is

1

y=0

1

z=0

(12 + y2 + z

2)(i)dzdy = i

1

y=0

z + y

2z +

1

3z

3

1

z=0

dy

= i

1

y=0

y

2 +4

3

dy = i

1

3y

3 +4

3y

1

0

=5

3i

The net contribution from the faces x = 0 and x = 1 is −23i + 5

3 i = i.Due to the symmetry of the scalar field u and the unit cube, the net contribution from the facesy = 0 and y = 1 is j while the net contribution from the faces z = 0 and z = 1 is k.

Adding, we obtain

S

udS = i + j + k

Key Point 4A scalar function integrated with respect to a normal vector dS gives a vector quantity.

When the surface does not lie in one of the planes Oxy, Oxz, Oyz, extra care must be taken whenfinding dS.

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Example 22

Find

S

(∇ · F )dS where F = 2xi + yzj + xyk and S is the surface of the

triangle with vertices at (0, 0, 0), (1, 0, 0) and (1, 1, 0).

Solution

Note that ∇ · F = 2 + z = 2 as z = 0 everywhere along S. As the triangle lies in the Oxy plane,the normal vector n = k and dS = kdydx.Thus,

S

(∇ · F )dS =

1

x=0

x

y=0

2dydxk =

1

0

2y

x

0

dxk =

1

0

2xdxk =

x21

0k = k

Here the scalar function being integrated was the divergence of a vector function.

Example 23

Find

S

f dS where f is the function 2x and S is the surface of the triangle

bounded by (0, 0, 0), (0, 1, 1) and (1, 0, 1). (See Figure 9.)

z

x

yArea

!3

2

1

2Area

Figure 9: The triangle defining the area S

HELM (2008):


39

Solution

The unit vector n is perpendicular to two vectors in the plane e.g. (j + k) and (i + k). The

vector (j + k) × (i + k) = i + j − k which has magnitude√

3. Hence the unit normal vectorn = 1√

3i + 1√

3j − 1√

3k.

As the area of the triangle S is√

32 and the area of its projection in the Oxy plane is 1

2 , the vector

dS =

√3/2

1/2n dydx = (i + j + k)dydx.

Thus

S

fdS = (i + j + k)

1

x=0

1−x

y=0

2x dydx

= (i + j + k)

1

x=0

2xy

1−x

y=0

dx

= (i + j + k)

1

x=0

(2x− 2x2)dx

= (i + j + k)

x

2 − 2

3x

3

1

0

=1

3(i + j + k)

Task


S

4x dS where S represents the trapezium with vertices

at (0, 0), (3, 0), (2, 1) and (0, 1).

(a) Find the vector dS:

Your solution

Answer

k dx dy

(b) Write the surface integral as a double integral:

Your solution

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Answer

It is easier to integrate first with respect to x. This gives

1

y=0

3−y

x=0

4x dx dy k.

The range of values of y is y = 0 to y = 1.

For each value of y, x varies from x = 0 to x = 3− y

(c) Evaluate this double integral:

Your solution

Answer38

3k

HELM (2008):


41

Exercises

1. Evaluate the integral

S

xydS where S is the triangle with vertices at (0, 0, 4), (0, 2, 0) and

(1, 0, 0).

2. Find the integral

S

xyzdS where S is the surface of the unit cube 0 ≤ x ≤ 1, 0 ≤ y ≤ 1,

0 ≤ z ≤ 1.


S

∇ · (x2

i + yzj + x2yk)

dS where S is the rectangle with vertices

at (1, 0, 0), (1, 1, 0), (1, 1, 1) and (1, 0, 1).

Answers 1.2

3i +

1

3j +

1

6k 2.

1

4(x + y + z), 3.

5

2i

Integrating a vector field

In a similar manner to the case of a scalar field, a vector field may be integrated over a surface.

Two common types of integral are

S

F (r) · dS and

S

F (r)× dS which integrate to a scalar and a

vector respectively. Again, when dS is expressed appropriately, the expression will reduce to a double

integral. The form

S

F (r) · dS has many important applications, e.g. the flux of a vector field such

as an electric or magnetic field.

Example 24


A

(x2yi + zj + (2x + y)k) · dS

over the area A where A is the square 0 ≤ x ≤ 1, 0 ≤ y ≤ 1, z = 0.

Solution

On A, the unit normal is dx dy k

∴

A

(x2yi + zj + (2x + y)k) · (k dx dy)

=

1

y=0

1

x=0

(2x + y) dx dy =

1

y=0

x

2 + xy

1

x=0

dy

=

1

y=0

(1 + y)dy =

y +

1

2y

2

1

0

=3

2

42 HELM (2008):


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Example 25

Evaluate

A

r · dS where A represents the surface of the unit cube

0 ≤ x ≤ 1, 0 ≤ y ≤ 1, 0 ≤ z ≤ 1 and r represents the vector xi + yj + zk .

Solution

The vector dS (in the direction of the normal vector) will be a constant vector on each face, butwill be different for each face.

On the face x = 0 , dS = −dy dz i and the integral on this face is 1

z=0

1

y=0

(0i + yj + zk) · (−dy dz i) =

1

z=0

1

y=0

0 dy dz = 0

Similarly on the face y = 0, dS = −dx dz j and the integral on this face is 1

z=0

1

x=0

(xi + 0j + zk) · (−dx dz j) =

1

z=0

1

x=0

0 dx dz = 0

Furthermore on the face z = 0, dS = −dx dy k and the integral on this face is 1

x=0

1

y=0

(xi + yj + 0k) · (−dx dy k) =

1

x=0

1

y=0

0 dx dy = 0

On these three faces, the contribution to the integral is thus zero.

However, on the face x = 1, dS = +dy dz i and the integral on this face is 1

z=0

1

y=0

(1i + yj + zk) · (+dy dz i) =

1

z=0

1

y=0

1 dy dz = 1

Similarly, on the face y = 1, dS = +dx dz j and the integral on this face is 1

z=0

1

x=0

(xi + 1j + zk) · (+dx dz j) =

1

z=0

1

x=0

1 dx dz = 1

Finally, on the face z = 1, dS = +dx dy k and the integral on this face is 1

y=0

1

x=0

(xi + yj + 1k) · (+dx dy k) =

1

y=0

1

x=0

1 dx dy = 1

Adding together the contributions gives

A

r · dS = 0 + 0 + 0 + 1 + 1 + 1 = 3

HELM (2008):


43


Magnetic flux

Introduction

The magnetic flux through a surface is given by

S

B·dS where S is the surface under consideration,

B is the magnetic field and dS is the vector normal to the surface.

Problem in words

The magnetic field generated by an infinitely long vertical wire on the z-axis, carrying a current I, isgiven by:

B =µ0I

2π

−yi + xj

x2 + y2

Find the flux through a rectangular region (with sides parallel to the axes) on the plane y = 0.


Find the integral

S

B · dS over the surface, x1 ≤ x ≤ x2, z1 ≤ z ≤ z2. (see Figure 10 which

shows part of the plane y = 0 for which the flux is to be found and a single magnetic field line. Thestrength of the field is inversely proportional to the distance from the axis.)

x

y

z

x1

x2

z1

z2

S

Figure 10: The surface S defined by x1 ≤ x ≤ x2, z1 ≤ z ≤ z2


On y = 0, B =µ0I

2πxj and dS = dx dz j so B · dS =

µ0I

2πxdx dz

The flux is given by the double integral: z2

z=z1

x2

x=x1

µ0I

2πxdx dz =

µ0I

2π

z2

z=z1

ln x

x2

x1

dz

=µ0I

2π

z2

z=z1

ln x2 − ln x1

dz

=µ0I

2π

zln x2 − ln x1

z2

z=z1

=µ0I

2π(z2 − z1) ln

x2

x1

44 HELM (2008):


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Interpretation

The magnetic flux increases in direct proportion to the extent of the side parallel to the axis (i.e.along the z-direction) but logarithmically with respect to the extent of the side perpendicular to theaxis (i.e. along the x-axis).

Example 26

If F = x2i + y2j + z2k, evaluate

S

F × dS where S is the part of the plane

z = 0 bounded by x = ±1, y = ±1.

Solution

Here dS = dx dy k and hence F × dS =

i j k

x2 y2 z2

0 0 dx dy

= y2 dx dy i− x2 dx dy j

S

F × dS =

1

y=−1

1

x=−1

y2dx dy i−

1

y=−1

1

x=−1

x2dx dy j

The first integral is 1

y=−1

1

x=−1

y2dx dy =

1

y=−1

y

2x

1

x=−1

dy =

1

y=−1

2y2dy =

2

3y

3

1

−1

=4

3

Similarly

1

y=−1

1

x=−1

x2dx dy =

4

3.

Thus

S

F × dS =4

3i− 4

3j

Key Point 5

(a) An integral of the form

S

F (r) · dS evaluates to a scalar.

(b) An integral of the form

S

F (r)× dS evaluates to a vector.

The vector function involved may be the gradient of a scalar or the curl of a vector.

HELM (2008):


45

Example 27

Integrate

S

(∇φ).dS where φ = x2 + 2yz and S is the area between y = 0 and

y = x2 for 0 ≤ x ≤ 1 and z = 0. (See Figure 11.)

x

y

1

1

S

Figure 11: The area S between y = 0 and y = x2, for 0 ≤ x ≤ 1 and z = 0

Solution

Here ∇φ = 2xi + 2zj + 2yk and dS = k dydx. Thus (∇φ).dS = 2ydydx and

S

(∇φ).dS =

1

x=0

x2

y=0

2y dydx

=

1

x=0

y

2

x2

y=0

dx =

1

x=0

x4dx

=

1

5x

5

1

0

=1

5

For integrals of the form

S

F · dS, non-Cartesian coordinates e.g. cylindrical polar or spherical

polar coordinates may be used. Once again, it is necessary to include any scale factors along withthe unit normal.

Example 28

Using cylindrical polar coordinates, (see 28.3), find the integral

S

F (r) ·dS

for F = ρzρ + z sin2φz and S being the complete surface (including ends) of the

cylinder ρ ≤ a, 0 ≤ z ≤ 1. (See Figure 12.)

z

x

y

! = a

z = 1

Figure 12: The cylinder ρ ≤ a, 0 ≤ z ≤ 1

46 HELM (2008):


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Solution

The integral

S

F (r) · dS must be evaluated separately for the curved surface and the ends.

For the curved surface, dS = ρadφdz (with the a coming from ρ the scale factor for φ and the factthat ρ = a on the curved surface.) Thus, F · dS = a2z dφdz and

S

F (r) · dS =

1

z=0

2π

φ=0

a2z dφdz

= 2πa2

1

z=0

z dz = 2πa2

1

2z

2

1

0

= πa2

On the bottom surface, z = 0 so F = 0 and the contribution to the integral is zero.On the top surface, z = 1 and dS = zρ dρdφ and F · dS = ρz sin2

φ dφdρ = ρ sin2φ dφdρ and

S

F (r) · dS =

a

ρ=0

2π

φ=0

ρ sin2φ dφdρ

= π

a

ρ=0

ρ dρ =1

2πa

2

So

S

F (r) · dS = πa2 +

1

2πa

2 =3

2πa

2


The current continuity equation

Introduction

When an electric current flows at a constant rate through a conductor, then the current continuityequation states that

S

J · dS = 0

where J is the current density (or current flow per unit area) and S is a closed surface. The equationis an expression of the fact that, under these conditions, the current flow into a closed volume equalsthe flow out.

Problem in words

A person is standing nearby when lightning strikes the ground. Find the potential difference betweenthe feet of that person.

HELM (2008):


47

Figure 13: Lightning: a current dissipating into the ground


The current from the lightning dissipates radially (see Fig 13).

(a) Find a relationship between the current I and current density J at a distance r from the

strike by integrating the current density over the hemisphere I =

S

J · dS

(b) Find the field E from the equation E =ρI

2π2rwhere E = |E| and I is the current.

(c) Find V from the integral

R2

R1

E · dr


Imagine a hemisphere of radius r level with the surface of the ground so that the point of lightningstrike is at its centre. By symmetry, the pattern of current flow from the point of strike will beuniform radial lines, and the magnitude of J will be a constant, i.e. over the curved surface of thehemisphere J = Jr.

Since the amount of current entering the hemisphere is I, then it follows that the current leavingmust be the same i.e.

I =

Sc

J · dS (where Sc is the curved surface of the hemisphere)

=

Sc

(Jr) · (dS r)

= J

Sc

dS

= 2πr2J [= surface area (2πr

2)× flux (J)]

since the surface area of a sphere is 4πr2. Therefore

J =I

2πr2

Note that if the current density J is uniformly radial over the curved surface, then so must be theelectric field E, i.e. E = Er. Using Ohm’s law

J = σE or E = ρJ

where σ = conductivity = 1/ρ.

48 HELM (2008):


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Hence E =ρI

2πr2

The potential difference between two points at radii R1 and R2 from the lightning strike is found byintegrating E between them, so that

V =

R2

R1

E · dr

=

R2

R1

E dr

=ρI

2π

R2

R1

dr

r2

=ρI

2π

−1

r

R2

R1

=ρI

2π

1

R1− 1

R2

=

ρI

2π

R2 −R1

R1R2

Interpretation

Suppose the lightning strength is a current I = 10, 000 A, that the person is 12 m away withfeet 0.35 m apart, and that the resistivity of the ground is 80 Ω m. Clearly, the worst case (i.e.maximum voltage) would occur when the difference between R1 and R2 is greatest, i.e. R1=12 mand R2=12.35 m which would be the case if both feet were on the same radial line. The voltageproduced between the person’s feet under these circumstances is

V =ρI

2π

1

R1− 1

R2

=80× 10000

2π

1

12− 1

12.35

≈ 300 V

Task

For F = (x2 + y2)i + (x2 + z2)j + 2xzk and S the square bounded by (1, 0, 1),

(1, 0,−1), (−1, 0,−1) and (−1, 0, 1) find the integral

S

F · dS

Your solution

Answer

dS = dxdzj

1

−1

1

−1

(x2 + z2) dxdz =

8

3

HELM (2008):


49

Task

For F = (x2 + y2)i + (x2 + z2)j + 2xzk and S being the rectangle bounded by(1, 0, 1), (1, 0,−1), (−1, 0,−1) and (−1, 0, 1) (i.e. the same F and S as in the

previous Task), find the integral

S

F × dS

Your solution

Answer 1

−1

1

−1

(−2xz)i +

1

−1

1

−1

(x2 + 0)k

dxdz =

4

3k

Exercises


S

∇φ · dS for φ = x2z sin y and S being the rectangle bounded by

(0, 0, 0), (1, 0, 1), (1, π, 1) and (0, π, 0).


S

(∇ × F ) × dS where F = xeyi + zeyj and S represents the unit

square 0 ≤ x ≤ 1, 0 ≤ y ≤ 1.

3. Using spherical polar coordinates (r, θ, φ), evaluate the integral

S

F ·dS where F = r cos θr

and S is the curved surface of the top half of the sphere r = a.

Answers 1. −2

3, 2. (e− 1)j, 3. πa3

2. Volume integrals involving vectors

Integrating a scalar function of a vector over a volume involves essentially the same procedure as in27.3. In 3D cartesian coordinates the volume element dV is dxdydz. The scalar function may

be the divergence of a vector function.

50 HELM (2008):


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Example 29

Integrate ∇ · F over the unit cube 0 ≤ x ≤ 1, 0 ≤ y ≤ 1, 0 ≤ z ≤ 1 where F isthe vector function x2yi + (x− z)j + 2xz2k.

Solution

∇ · F =∂

∂x(x2y) +

∂

∂y(x− z) +

∂

∂z(2xz2) = 2xy + 4xz

The integral is

1

x=0

1

y=0

1

z=0

(2xy + 4xz)dzdydx =

1

x=0

1

y=0

2xyz + 2xz

2

1

0

dydx

=

1

x=0

1

y=0

(2xy + 2x) dydx =

1

x=0

xy

2 + 2xy

1

0

dx

=

1

x=0

3xdx =

3

2x

2

1

0

=3

2

Key Point 6The volume integral of a scalar function (including the divergence of a vector) is a scalar.

Task

Using spherical polar coordinates (r, θ, φ) and the vector field F = r2 r+r2 sin θ θ,

evaluate the integral

V

∇ · F dV over the sphere given by 0 ≤ r ≤ a.

Your solution

Answer

∇ · F = 4r + 2r cos θ,

a

r=0

π

θ=0

2π

φ=0

(4r + 2r cos θ)r2 sin θ dφdθdr = 4πa4

The r2 sin θ term comes from the Jacobian for the transformation from spherical to cartesian coor-dinates (see 27.4 and 28.3).

HELM (2008):


51

Exercises

1. Evaluate

V

∇ · FdV when F is the vector field yzi + xyj and V is the unit cube

0 ≤ x ≤ 1, 0 ≤ y ≤ 1, 0 ≤ z ≤ 1

2. For the vector field F = (x2y+sin z)i+(xy2+ez)j+(z2+xy)k, find the integral

V

∇·FdV

where V is the volume inside the tetrahedron bounded by x = 0, y = 0, z = 0 and x+y+z = 1.

Answers 1. ∇ · F = x,1

22.

7

60

Integrating a vector function over a volume integral is similar, but less common. Care should betaken with the various components. It may help to think in terms of a separate volume integral foreach component. The vector function may be of the form ∇f or ∇× F .

Example 30

Integrate the function F = x2i+2j over the prism given by 0 ≤ x ≤ 1, 0 ≤ y ≤ 2,0 ≤ z ≤ (1− x). (See Figure 14.)

z

x

y

1 2

1

Figure 14: The prism bounded by 0 ≤ x ≤ 1, 0 ≤ y ≤ 2, 0 ≤ z ≤ (1− x)

Solution

The integral is

1

x=0

2

y=0

1−x

z=0

(x2i + 2j)dzdydx =

1

x=0

2

y=0

x

2zi + 2zj

1−x

z=0

dydx

=

1

x=0

2

y=0

x

2(1− x)i + 2(1− x)j

dydx =

1

x=0

2

y=0

(x2 − x

3)i + (2− 2x)j

dydx

=

1

x=0

(2x2 − 2x3)i + (4− 4x)j

dx =

(2

3x

3 − 1

2x

4)i + (4x− 2x2)j

1

0

=1

6i + 2j

52 HELM (2008):


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Example 31

For F = x2yi + y2j evaluate

V

(∇×F )dV where V is the volume under the

plane z = x + y + 2 (and above z = 0) for −1 ≤ x ≤ 1, −1 ≤ y ≤ 1.

Solution

∇× F =

i j k

∂

∂x

∂

∂y

∂

∂z

x2y y2 0

= −x2k

so

V

(∇× F )dV =

1

x=−1

1

y=−1

x+y+2

z=0

(−x2)k dzdydx

=

1

x=−1

1

y=−1

(−x

2)zk

x+y+2

z=0

dydx

=

1

x=−1

1

y=−1

−x

3 − x2y − 2x2

dydx k

=

1

x=−1

−x

3y − 1

2x

2y

2 − 2x2y

1

y=−1

dx k

=

1

x=−1

−2x3 − 0− 4x2

dx k =

−1

2x

4 − 4

3x

3

1

−1

k = −8

3k

(!1,!1, 0)

(!1, 1, 0)

(1, 1, 4)

(1, 1, 0)

(1,!1, 0)

z

x

y

Figure 15: The plane defined by z = x + y + z, for z > 0, −1 ≤ x ≤ 1, −1 ≤ y ≤ 1

HELM (2008):


53

Key Point 7The volume integral of a vector function (including the gradient of a scalar or the curl of a vector)is a vector.

Task


V

FdV for the case where F = xi + y2j + zk and V is the

cube −1 ≤ x ≤ 1, −1 ≤ y ≤ 1, −1 ≤ z ≤ 1.

Your solution

Answer 1

x=−1

1

y=−1

1

z=−1

(xi + y2j + zk)dzdydx =

8

3j

Exercises

1. For f = x2 + yz, and V the volume bounded by y = 0, x + y = 1 and −x + y = 1 for

−1 ≤ z ≤ 1, find the integral

V

(∇f)dV .


V

(∇× F )dV for the case where F = xzi + (x3 + y3)j − 4yk and V is

the cube −1 ≤ x ≤ 1, −1 ≤ y ≤ 1, −1 ≤ z ≤ 1.

Answers

1.

V

(2xi + zj + yk)dV =2

3k,

2.

V

(−4i + xj + 3x2k)dV = −32i + 8k

54 HELM (2008):


Line Integrals

29.1

Introduction

workbook 28 considered the differentiation of scalar and vector fields. Here we considerhow to integrate such fields along a line. Firstly, integrals involving scalars along a line will beconsidered. Subsequently, line integrals involving vectors will be considered. These can give scalaror vector answers depending on the form of integral involved. Of particular interest are the integralsof conservative vector fields.

Prerequisites


• have a thorough understanding of the basictechniques of integration

• be familiar with the operators div, grad andcurl

Learning Outcomes


• integrate a scalar or vector quantity along aline

2 HELM (2008):


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1. Line integrals

28 was concerned with evaluating an integral over all points within a rectangle or other shape(or over a cuboid or other volume). In a related manner, an integral can take place over a line orcurve running through a two-dimensional (or three-dimensional) region. Line integrals may involvescalar or vector fields. Those involving scalar fields are dealt with first.

Line integrals in two dimensions

A line integral in two dimensions may be written as

C

F (x, y)dw

There are three main features determining this integral:

F (x, y): This is the scalar function to be integrated e.g. F (x, y) = x2 + 4y2.

C: This is the curve along which integration takes place. e.g. y = x2 or x = sin y

or x = t− 1; y = t2 (where x and y are expressed in terms of a parameter t).

dw: This gives the variable of the integration. Three main cases are dx, dy and ds.Here ‘s’ is arc length and so indicates position along the curve C.

ds may be written as ds =

(dx)2 + (dy)2 or ds =

1 +

dy

dx

2

dx.

A fourth case is when F (x, y) dw has the form: F1dx+F2dy. This is a combinationof the cases dx and dy.

The integral

C

F (x, y) ds represents the area beneath the surface z = F (x, y) but above the curve

C.

The integrals

C

F (x, y) dx and

C

F (x, y) dy represent the projections of this area onto the xz

and yz planes respectively.

A particular case of the integral

C

F (x, y) ds is the integral

C

1 ds. This is a means of calculating

the length along a curve i.e. an arc length.

!

C

f(x, y)dy

!

C

f(x, y)dx

!

C

f(x, y)ds

curve C

x

y

z

Figure 1: Representation of a line integral and its projections onto the xz and yz planes

HELM (2008):

Section 29.1: Line Integrals

3

The technique for evaluating a line integral is to express all quantities in the integral in terms of asingle variable. If the integral is with respect to ’x’ or ’y’, then the curve ’C’ andthe function ’F ’ may be expressed in terms of the relevant variable. If the integral is withrespect to ds, normally all quantities are expressed in terms of x. If x and y are given in terms of aparameter t, then t is used as the variable.

Example 1

Find

c

x (1 + 4y) dx where C is the curve y = x2, starting from x = 0, y = 0

and ending at x = 1, y = 1.

Solution

As this integral concerns only points along C and the integration is carried out with respect to x,y may be replaced by x2. The limits on x will be 0 to 1. So the integral becomes

C

x(1 + 4y) dx =

1

x=0

x1 + 4x2

dx =

1

x=0

x + 4x3

dx

=

x2

2+ x

4

1

0

=

1

2+ 1

− (0) =

3

2

Example 2

Find

c

x (1 + 4y) dy where C is the curve y = x2, starting from

x = 0, y = 0 and ending at x = 1, y = 1. This is the same as Example 1 otherthan dx being replaced by dy.

Solution

As this integral concerns only points along C and the integration is carried out with respect to y,everything may be expressed in terms of y, i.e. x may be replaced by y1/2. The limits on y willbe 0 to 1. So the integral becomes

C

x(1 + 4y) dy =

1

y=0

y1/2 (1 + 4y) dx =

1

y=0

y

1/2 + 4y3/2dx

=

2

3y

3/2 +8

5x

5/2

1

0

=

2

3+

8

5

− (0) =

34

15

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Example 3

Find

c

x (1 + 4y) ds where C is the curve y = x2, starting from x = 0, y = 0

and ending at x = 1, y = 1. This is the same integral and curve as the previoustwo examples but the integration is now carried out with respect to s, the arclength parameter.

Solution

As this integral is with respect to x, all parts of the integral can be expressed in terms of x, Along

y = x2, ds =

1 +

dy

dx

2

dx =

1 + (2x)2dx =

√1 + 4x2dx

So, the integral is

c

x (1 + 4y) ds =

1

x=0

x1 + 4x2

√1 + 4x2 dx =

1

x=0

x1 + 4x2

3/2dx

This can be evaluated using the transformation u = 1 + 4x2 so du = 8xdx i.e. x dx =du

8.

When x = 0, u = 1 and when x = 1, u = 5.Hence,

1

x=0

x1 + 4x2

3/2dx =

1

8

5

u=1

u3/2

du

=1

8× 2

5

u

5/2

5

1

=1

20

55/2 − 1

≈ 2.745

Note that the results for Examples 1,2 and 3 are all different: Example 3 is the area between a curveand a surface above; Examples 1 and 2 give projections of this area onto other planes.

Example 4

Find

C

xy dx where, on C, x and y are given in terms of a parameter t by

x = 3t2, y = t3 − 1 for t varying from 0 to 1.

Solution

Everything can be expressed in terms of t, the parameter. Here x = 3t2 so dx = 6t dt. The limitson t are t = 0 and t = 1. The integral becomes

C

xy dx =

1

t=0

3t2 (t3 − 1) 6t dt =

1

t=0

(18t6 − 18t3) dt

=

18

7t7 − 18

4t4

1

0

=18

7− 9

2− 0 = −27

14

HELM (2008):


5

Key Point 1A line integral is normally evaluated by expressing all variables in terms of one variable.

In general

C

f(x, y) ds =

C

f(x, y) dy =

C

f(x, y) dx

Task

For F (x, y) = 2x + y2, find (i)

C

F (x, y) dx, (ii)

C

F (x, y) dy,

(iii)

C

F (x, y) ds where C is the line y = 2x from (0, 0) to (1, 2).

Express each integral as a simple integral with respect to a single variable and hence evaluate eachintegral:

Your solution

Answer

(i)

1

x=0

(2x + 4x2) dx =7

3, (ii)

2

y=0

(y + y2) dy =

14

3, (iii)

1

x=0

(2x + 4x2)√

5 dx =7

3

√5

6 HELM (2008):


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Task

Find (i)

C

F (x, y) dx, (ii)

C

F (x, y) dy, (iii)

C

F (x, y) ds where F (x, y) = 1

and C is the curve y = 12x

2 − 14 ln x from (1, 1

2) to (2, 2− 14 ln 2).

Your solution

Answer

(i)

2

1

1 dx = 1, (ii)

2−(1/4) ln 2

1/2

1 dy =3

2− 1

4ln 2, (iii) y =

1

2x

2 − 1

4ln x⇒ dy

dx= x− 1

4x

∴

1 ds =

2

1

1 + (x− 1

4x)2 dx =

2

1

x2 +

1

2+

1

16x2dx =

2

1

(x +1

4x) dx =

3

2+

1

4ln 2.

Task

Find (i)

C

F (x, y) dx, (ii)

C

F (x, y) dy, (iii)

C

F (x, y) ds

where F (x, y) = sin 2x and C is the curve y = sin x from (0, 0) to (π

2, 1).

Your solution

Answer

(i)

π/2

0

sin 2x dx = 1, (ii)

π/2

0

2 sin x cos2x dx =

2

3

(iii)

π/2

0

sin 2x√

1 + cos2 x dx =2

3(2√

2− 1), using the substitution u = 1 + cos2 x.

HELM (2008):


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2. Line integrals of scalar products

Integrals of the form

C

F · dr occur in applications such as the following.

!r T

B

A

v

S (current position)dr

Figure 2: Schematic for cyclist travelling from A to B into a head wind

Consider a cyclist riding along the road from A to B (Figure 2). Suppose it is necessary to find thetotal work the cyclist has to do in overcoming a wind of velocity v.

On moving from S to T , along an element δr of road, the work done is given by ‘Force × distance’= |F |× |δr| cos θ where F , the force, is directly proportional to v, but in the opposite direction, and|δr| cos θ is the component of the distance travelled in the direction of the wind.So, the work done travelling δr is −kv · δr. Letting δr become infinitesimally small, the work done

becomes −kv · dr and the total work is −k

B

A

v · dr.

This is an example of the integral along a line, of the scalar product of a vector field, with a vectorelement of the line. The term scalar line integral is often used for integrals of this form. Thevector dr may be considered to be dx i + dy j + dz k.

Multiplying out the scalar product, the ’scalar line integral’ of the vector F along contour C, is given

by

C

F · dr and equals

C

Fx dx + Fy dy + Fz dz in three dimensions, and

C

Fx dx + Fy dyin two dimensions, where Fx, Fy, Fz are the components of F .

If the contour C has its start and end points in the same positions i.e. it represents a closed contour,

the symbol

C

rather than

C

is used, i.e.

C

F · dr .

As before, to evaluate the line integral, express the path and the function F in terms of either x, y

and z, or in terms of a parameter t. Note that t often represents time.

Example 5

Find

C

2xy dx− 5x dy where C is the curve y = x3 0 ≤ x ≤ 1.

[This is the integral

C

F · dr where F = 2xyi− 5xj and dr = dx i + dy j.]

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Solution

It is possible to split this integral into two different integrals and express the first term as a functionof x and the second term as a function of y. However, it is also possible to express everything interms of x. Note that on C, y = x3 so dy = 3x2 dx and the integral becomes

C

2xy dx− 5x dy =

1

x=0

2x x

3dx− 5x 3x2

dx

=

1

0

(2x4 − 15x3) dx

=

2

5x

5 − 15

4x

4

1

0

=2

5− 15

4− 0 = −67

20

Key Point 2

An integral of the form

C

F · dr may be expressed as

C

Fx dx + Fy dy + Fz dz. Knowing the

expression for the path C, every term in the integral can be further expressed in terms of one ofthe variables x, y or z or in terms of a parameter t and hence integrated.

If an integral is two-dimensional there are no terms involving z.

The integral

C

F · dr evaluates to a scalar.

Example 6

Three paths from (0, 0) to (1, 2) are defined by

(a) C1 : y = 2x(b) C2 : y = 2x2

(c) C3 : y = 0 from (0, 0) to (1, 0) and x = 1 from (1, 0) to (1, 2)

Sketch each path, and along each path find

F · dr, where F = y2i + xyj.

HELM (2008):


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Solution

(a)

F · dr =

y

2dx + xydy

. Along y = 2x,

dy

dx= 2 so dy = 2dx. Then

C1

F · dr =

1

x=0

(2x)2

dx + x (2x) (2dx)

=

1

0

4x2 + 4x2

dx =

1

0

8x2dx =

8

3x

2

1

0

=8

3

y = 2x

C1

A(1, 2)

x

y

1

2

Figure 3(a): Integration along path C1

(b)

F · dr =

y

2dx + xydy

. Along y = 2x2,

dy

dx= 4x so dy = 4xdx. Then

C2

F · dr =

1

x=0

2x2

2dx + x

2x2

(4xdx)

=

1

0

12x4dx =

12

5x

5

1

0

=12

5

y = 2x2

A(1, 2)

C2

y

1

2

x

Figure 3(b): Integration along path C2

Note that the answer is different to part (a), i.e., the line integral depends upon the path taken.

(c) As the contour C3, has two distinct parts with different equations, it is necessary to break thefull contour OA into the two parts, namely OB and BA where B is the point (1, 0). Hence

C3

F · dr =

B

O

F · dr +

A

B

F · dr

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Solution (contd.)

Along OB, y = 0 so dy = 0. Then

B

O

F · dr =

1

x=0

02

dx + x× 0× 0

=

1

0

0dx = 0

Along AB, x = 1 so dx = 0. Then

B

A

F · dr =

2

y=0

y

2 × 0 + 1× y × dy

=

2

0

ydy =

1

2y

2

2

0

= 2.

Hence

C3

F · dr = 0 + 2 = 2

y

1

2

y = 0

x = 1

C3

A(1, 2)

xOB

Figure 3(c): Integration along path C3

Once again, the result is path dependent.

Key Point 3In general, the value of a line integral depends on the path of integration as well as upon the endpoints.

HELM (2008):


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Example 7

Find

O

A

F · dr, where F = y2i+xyj (as in Example 6) and the path C4 from A

to O is the straight line from (1, 2) to (0, 0), that is the reverse of C1 in Example6(a).

Deduce

C

F · dr, the integral around the closed path C formed by the parabola

y = 2x2 from (0, 0) to (1, 2) and the line y = 2x from (1, 2) to (0, 0).

Solution

Reversing the path interchanges the limits of integration, which results in a change of sign for thevalue of the integral.

O

A

F · dr = −

A

O

F · dr = −8

3

The integral along the parabola (calculated in Example 6(b)) evaluates to12

5, then

C

F · dr =

C2

F · dr +

C4

F · dr =12

5− 8

3= − 4

15≈ −0.267

Example 8

Consider the vector field

F = y2z

3i + 2xyz

3j + 3xy

2z

2k

Let C1 and C2 be the curves from O = (0, 0, 0) to A = (1, 1, 1), given by

C1 : x = t, y = t, z = t (0 ≤ t ≤ 1)

C2 : x = t2, y = t, z = t

2 (0 ≤ t ≤ 1)

(a) Evaluate the scalar integral of the vector field along each path.

(b) Find the value of

C

F · dr where C is the closed path along C1 from

O to A and back along C2 from A to O.

12 HELM (2008):


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Solution

(a) The path C1 is given in terms of the parameter t by x = t, y = t and z = t. Hence

dx

dt=

dy

dt=

dz

dt= 1 and

dr

dt=

dx

dti +

dy

dtj +

dz

dtk = i + j + k

Now by substituting for x = y = z = t in F we have

F = t5i + 2t5j + 3t5k

Hence F · dr

dt= t

5 + 2t5 + 3t5 = 6t5. The values of t = 0 and t = 1 correspond to the

start and end point of C1 and so these are the required limits of integration. Now

C1

F · dr =

1

0

F · dr

dtdt =

1

0

6t5dt =

t6

1

0

= 1

For the path C2 the parameterisation is x = t2, y = t and z = t2 sodr

dt= 2ti+ j +2tk.

Substituting x = t2, y = t and z = t2 in F we have

F = t8i + 2t9j + 3t8k and F · dr

dt= 2t9 + 2t9 + 6t9 = 10t9

C2

F · dr =

1

0

10t9dt =

t10

1

0

= 1

(b) For the closed path C

C

F · dr =

C1

F · dr −

C2

F · dr = 1− 1 = 0

(Note: A line integral round a closed path is not necessarily zero - see Example 7.)

Further points on Example 8

Vector Field Path Line IntegralF C1 1F C2 1F closed 0

Note that the value of the line integral of F is 1 for both paths C1 and C2. In fact, this result wouldhold for any path from (0, 0, 0) to (1, 1, 1).

The field F is an example of a conservative vector field; these are discussed in detail in thenext subsection.

In

C

F · dr, the vector field F may be the gradient of a scalar field or the curl of a vector field.

HELM (2008):


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Task

Consider the vector field

G = xi + (4x− y)j

Let C1 and C2 be the curves from O = (0, 0, 0) to A = (1, 1, 1), given by

C1 : x = t, y = t, z = t (0 ≤ t ≤ 1)

C2 : x = t2, y = t, z = t

2 (0 ≤ t ≤ 1)

(a) Evaluate the scalar integral

C

G · dr of each vector field along each

path.

(b) Find the value of

C

G · dr where C is the closed path along C1 from

O to A and back along C2 from A to O.

Your solution

14 HELM (2008):


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Answer

(a) The path C1 is given in terms of the parameter t by x = t, y = t and z = t. Hence

dx

dt=

dy

dt=

dz

dt= 1 and

dr

dt=

dx

dti +

dy

dtj +

dz

dtk = i + j + k

Substituting for x = y = z = t in G we have

G = ti + 3tj and G · dr

dt= t + 3t = 4t

The limits of integration are t = 0 and t = 1, then

C1

G · dr =

1

0

G · dr

dtdt =

1

0

4tdt =

2t2

1

0

= 2

For the path C2 the parameterisation is x = t2, y = t and z = t2 sodr

dt= 2ti+ j +2tk.

Substituting x = t2, y = t and z = t2 in G we have

G = t2i +

4t2 − t

j and G · dr

dt= 2t3 + 4t2 − t

C2

G · dr =

1

0

2t3 + 4t2 − t

dt =

1

2t4 +

4

3t3 − 1

2t2

1

0

=4

3

(b) For the closed path C

C

G · dr =

C1

G · dr −

C2

G · dr = 2− 4

3=

2

3

(Note: The value of the integral around the closed path is non-zero, unlike Example 8.)

Example 9

Find

C

∇(x2

y)· dr where C is the contour y = 2x− x2 from (0, 0) to (2, 0).

Here, ∇ refers to the gradient operator, i.e. ∇φ ≡ grad φ

Solution

Note that ∇(x2y) = 2xyi + x2j so the integral is

C

2xy dx + x

2dy

.

On y = 2x− x2, dy = (2− 2x) dx so the integral becomes

C

2xy dx + x

2dy

=

2

x=0

2x(2x− x

2) dx + x2(2− 2x) dx

=

2

0

(6x2 − 4x3) dx =

2x3 − x

4

2

0

= 0

HELM (2008):


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Example 10

Two paths from (0, 0) to (4, 2) are defined by

(a) C1 : y =1

2x 0 ≤ x ≤ 4

(b) C2 : The straight line y = 0 from (0, 0) to (4, 0) followed byC3 : The straight line x = 4 from (4, 0) to (4, 2)

For each path find

C

F · dr, where F = 2xi + 2yj.

Solution

(a) For the straight line y =1

2x we have dy =

1

2dx

Then,

C1

F · dr =

C1

2x dx + 2y dy =

4

0

2x +

x

2

dx =

4

0

5x

2dx = 20

(b) For the straight line from (0, 0) to (4, 0) we have

C2

F · dr =

4

0

2x dx = 16

For the straight line from (4, 0) to (4, 2) we have

C3

F · dr =

2

0

2y dy = 4

Adding these two results gives

C

F · dr = 16 + 4 = 20

Task

Evaluate

C

F · dr, where F = (x − y)i + (x + y)j along each of the following

paths

(a) C1 : from (1, 1) to (2, 4) along the straight line y = 3x− 2:

(b) C2 : from (1, 1) to (2, 4) along the parabola y = x2:

(c) C3 : along the straight line x = 1 from (1, 1) to (1, 4) then along thestraight line y = 4 from (1, 4) to (2, 4).

Your solution

16 HELM (2008):


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Answer

(a)

2

1

(10x− 4) dx = 11,

(b)

2

1

(x + x2 + 2x3) dx =

34

3, (this differs from (a) showing path dependence)

(c)

4

1

(1 + y) dy +

2

1

(x− 4) dx = 8

Task

For the function F and paths in the last Task, deduce

F · dr for the closed

paths

(a) C1 followed by the reverse of C2.

(b) C2 followed by the reverse of C3.

(c) C3 followed by the reverse of C1.

Your solution

Answer

(a) −1

3, (b)

10

3, (c) −3. (note that all these are non-zero.)

HELM (2008):


17

Exercises

1. Consider

C

F · dr, where F = 3x2y2i + (2x3y− 1)j. Find the value of the line integral along

each of the paths from (0, 0) to (1, 4).

(a) y = 4x (b) y = 4x2 (c) y = 4x1/2 (d) y = 4x3

2. Consider the vector field F = 2xi + (xz− 2)j + xyk and the two curves between (0, 0, 0) and(1,−1, 2) defined by

C1 : x = t2, y = −t, z = 2t for 0 ≤ t ≤ 1.C2 : x = t− 1, y = 1− t, z = 2t− 2 for 1 ≤ t ≤ 2.

(a) Find

C1

F · dr,

C2

F · dr

(b) Find

C

F · dr where C is the closed path from (0, 0, 0) to (1,−1, 2) along C1 and back

to (0, 0, 0) along C2.

3. Consider the vector field G = x2zi + y2zj + 13(x

3 + y3)k and the two curves between (0, 0, 0)and (1,−1, 2) defined by

C1 : x = t2, y = −t, z = 2t for 0 ≤ t ≤ 1.C2 : x = t− 1, y = 1− t, z = 2t− 2 for 1 ≤ t ≤ 2.

(a) Find

C1

G · dr,

C2

G · dr

(b) Find

C

G · dr where C is the closed path from (0, 0, 0) to (1,−1, 2) along C1 and back

to (0, 0, 0) along C2.

4. Find

C

F · dr) along y = 2x from (0, 0) to (2, 4) for

(a) F = ∇(x2y)

(b) F = ∇× (12x

2y2k) [Here ∇× f represents the curl of f ]

Answers

1. All are 12, and in fact the integral would be 12 for any path from (0,0) to (1,4).

2 (a) 2, 53 (b) 1

3 .

3 (a) 0, 0 (b) 0.

4. (a)

C

2xy dx + x2

dy = 16, (b)

C

x2y dx− xy

2dy = −24.

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3. Conservative vector fields

For some line integrals in the previous section, the value of the integral depended only on the vectorfield F and the start and end points of the line but not on the actual path between the start andend points. However, for other line integrals, the result depended on the actual details of the pathof the line.

Vector fields are classified according to whether the line integrals are path dependent or path indepen-dent. Those vector fields for which all line integrals between all pairs of points are path independentare called conservative vector fields.

There are five properties of a conservative vector field (P1 to P5 below). It is impossible to check thevalue of every line integral over every path, but it is possible to use any one of these five properties(particularly property P3 below) to determine whether or not a vector field is conservative. Theseproperties are also used to simplify calculations with conservative vector fields over non-closed paths.

P1 The line integral

B

A

F · dr depends only on the end points A and B and is independent of

the actual path taken.

P2 The line integral around any closed curve is zero. That is

C

F · dr = 0 for all C.

P3 The curl of a conservative vector field F is zero i.e. ∇× F = 0.

P4 For any conservative vector field F , it is possible to find a scalar field φ such that ∇φ = F .

Then,

C

F · dr = φ(B)− φ(A) where A and B are the start and end points of contour C.

[This is sometimes called the Fundamental Theorem of Line Integrals and is comparable withthe Fundamental Theorem of Calculus.]

P5 All gradient fields are conservative. That is, F = ∇φ is a conservative vector field for anyscalar field φ.

Example 11

Consider the following vector fields.1. F 1 = y2i + xyj (Example 6) 2. F 2 = 2xi + 2yj (Example 10)

3. F 3 = y2z3i + 2xyz3j + 3xy2z2k (Example 8)

4. F 4 = xi + (4x− y) j (Task on page 14)

Determine which of these vector fields are conservative where possible by referringto the answers given in the solution. For those that are conservative find a scalarfield φ such that F = ∇φ and use property P4 to verify the values of the lineintegrals.

Solution

1. Two different values were obtained for line integrals over the paths C1 and C2. Hence, by P1,F 1 is not conservative. [It is also possible to reach this conclusion from P3 by finding that∇× F = −yk = 0.]

HELM (2008):


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Solution (contd.)

2. For the closed path consisting of C2 and C3 from (0, 0) to (4, 2) and back to (0, 0) along C1

we obtain the value 20 + (−20) = 0. This alone does not mean that F 2 is conservative as therecould be other paths giving different values. So by using P3

∇× F 2 =

i j k

∂

∂x

∂

∂y

∂

∂z

2x 2y 0

= i(0− 0)− j(0− 0) + k(0− 0) = 0

As ∇× F 2 = 0, P3 gives that F 2 is a conservative vector field.

Now, find a φ such that F 2 = ∇φ. Then∂φ

∂xi +

∂φ

∂yj = 2xi + 2yj.

Thus

∂φ

∂x= 2x ⇒ φ = x2 + f(y)

∂φ

∂y= 2y ⇒ φ = y2 + g(x)

⇒ φ = x

2 + y2(+ constant)

Using P4:

(4,2)

(0,0)

F 2 · dr =

(4,2)

(0,0)

(∇φ) · dr = φ(4, 2)− φ(0, 0) = (42 + 22)− (02 + 02) = 20.

3. The fact that line integrals along two different paths between the same start and end pointshave the same value is consistent with F 3 being a conservative field according to P1. So too is thefact that the integral around a closed path is zero according to P2. However, neither fact can beused to conclude that F 3 is a conservative field. This can be done by showing that ∇× F 3 = 0.

Now,

i j k

∂

∂x

∂

∂y

∂

∂z

y2z3 2xyz3 3xy2z2

= (6xyz2 − 6xyz2)i− (3y2z2 − 3y2z2)j + (2yz3 − 2yz3)k = 0.

As ∇× F 3 = 0, P3 gives that F 3 is a conservative field.

To find φ that satisfies ∇φ = F 3, it is necessary to satisfy

∂φ

∂x= y2z3 → φ = xy2z3 + f(y, z)

∂φ

∂y= 2xyz3 → φ = xy2z3 + g(x, z)

∂φ

∂z= 3xy2z2 → φ = xy2z3 + h(x, y)

→ φ = xy2z3

Using P4:

(1,1,1)

(0,0,0)

F 3 · dr = φ(1, 1, 1)− φ(0, 0, 0) = 1− 0 = 1 in agreement with Example 8(a).

20 HELM (2008):


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Solution (contd.)

4. As the integral along C1 is 2 and the integral along C2 (same start and end points but differentintermediate points) is 4

3 , F4 is not a conservative field using P1.Note that ∇ × F 4 = 4k = 0 so, using P3, this is an independent conclusion that F 4 is notconservative.


Work done moving a charge in an electric field

Introduction

If a charge, q, is moved through an electric field, E, from A to B, then the workrequired is given by the line integral

WAB = −q

B

A

E · dr

Problem in words

Compare the work done in moving a charge through the electric field around a point charge in avacuum via two different paths.


An electric field E is given by

E =Q

4πε0r2r

=Q

4πε0(x2 + y2 + z2)×

xi + yj + zk

x2 + y2 + z2

=Q(xi + yj + zk)

4πε0(x2 + y2 + z2)32

where r is the position vector with magnitude r and unit vector r, and1

4π0is a combination of

constants of proportionality, where 0 = 10−9/36π F m−1.

Given that Q = 10−8C, find the work done in bringing a charge of q = 10−10C from the pointA = (10, 10, 0) to the point B = (1, 1, 0) (where the dimensions are in metres)

(a) by the direct straight line y = x, z = 0

(b) by the straight line pair via C = (10, 1, 0)

HELM (2008):


21

O

A

B Cx

y

ab

b

Figure 4: Two routes (a and b) along which a charge can move through an electric field

The path comprises two straight lines from A = (10, 10, 0) to B = (1, 1, 0) via C = (10, 1, 0) (seeFigure 4).


(a) Here Q/(4πε0) = 90 so

E =90[xi + yj]

(x2 + y2)32

as z = 0 over the region of interest. The work done

WAB = −q

B

A

E · dr

= −10−10

B

A

90

(x2 + y2)32

[xi + yj] · [dxi + dyj]

Using y = x, dy = dx

WAB = −10−10

1

x=10

90

(2x2)32

x dx + x dx

= −10−10

1

10

90

(2√

2)x−3 2x dx

=90×−10−10

√2

1

10

x−2

dx

=9×−10−9

√2

− x

−1

1

10

=9× 10−9

√2

x−1

1

10

=9× 10−9

√2

[1− 0.1]

= 5.73× 10−9 J

22 HELM (2008):


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(b) The first part of the path is A to C where x = 10, dx = 0 and y goes from 10 to 1.

WAC = −q

C

A

E · dr

= −10−10

1

y=10

90

(100 + y2)32

[xi + yj] · [0i + dyj]

= −10−10

1

10

90y dy

(100 + y2)32

= −10−10

101

u=200

45 du

u32

(substituting u = 100 + y2, du = 2y dy)

= −45× 10−10

101

200

u− 3

2 du

= −45× 10−10−2u−

12

101

200

= 45× 10−10

2√101

− 2√200

= 2.59× 10−10J

The second part is C to B, where y = 1, dy = 0 and x goes from 10 to 1.

WCB = −10−10

1

x=10

90

(x2 + 1)32

[xi + yj] · [dxi + 0j]

= −10−10

1

10

90x dx

(x2 + 1)32

= −10−10

2

u=101

45 du

u32

(substituting u = x2 + 1, du = 2x dx)

= −45× 10−10

2

101

u− 3

2 du

= −45× 10−10−2u−

12

2

101

= 45× 10−10

2√2− 2√

101

= 5.468× 10−9J

The sum of the two components WAC and WCB is 5.73× 10−9J.

Therefore the work done over the two paths (a) and (b) is identical.

Interpretation

In fact, the work done is independent of the route taken as the electric field E around a point chargein a vacuum is a conservative field.

HELM (2008):


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Example 12

1. Show that I =

(2,1)

(0,0)

(2xy + 1)dx + (x2 − 2y)dy

is independent of the

path taken.

2. Find I using property P1. (Page 19)

3. Find I using property P4. (Page 19)

4. Find I =

C

(2xy + 1)dx + (x2 − 2y)dy

where C is

(a) the circle x2 + y2 = 1

(b) the square with vertices (0, 0), (1, 0), (1, 1), (0, 1).

Solution

1. The integral I =

(2,1)

(0,0)

(2xy + 1)dx + (x2 − 2y)dy

may be re-written

C

F · dr where

F = (2xy + 1)i + (x2 − 2y)j.

Now ∇× F =

i j k

∂

∂x

∂

∂y

∂

∂z

2xy + 1 x2 − 2y 0

= 0i + 0j + 0k = 0

As ∇×F = 0, F is a conservative field and I is independent of the path taken between (0, 0)and (2, 1).

2. As I is independent of the path taken from (0, 0) to (2, 1), it can be evaluated along anysuch path. One possibility is the straight line y = 1

2x. On this line, dy = 12dx. The integral

I becomes

I =

(2,1)

(0,0)

(2xy + 1)dx + (x2 − 2y)dy

=

2

x=0

(2x× 1

2x + 1)dx + (x2 − x)

1

2dx

=

2

0

(3

2x

2 − 1

2x + 1)dx

=

1

2x

3 − 1

4x

2 + x

2

0

= 4− 1 + 2− 0 = 5

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Solution (contd.)

3. If F = ∇φ then

∂φ

∂x= 2xy + 1 → φ = x2y + x + f(y)

∂φ

∂y= x2 − 2y → φ = x2y − y2 + g(x)

→ φ = x2y + x− y2 + C.

These are consistent if φ = x2y + x − y2 (plus a constant which may be omitted since itcancels).So I = φ(2, 1)− φ(0, 0) = (4 + 2− 1)− 0 = 5

4. As F is a conservative field, all integrals around a closed contour are zero.

Exercises

1. Determine whether the following vector fields are conservative

(a) F = (x− y)i + (x + y)j

(b) F = 3x2y2i + (2x3y − 1)j

(c) F = 2xi + (xz − 2)j + xyk

(d) F = x2zi + y2zj + 13(x

3 + y3)k

2. Consider the integral

C

F · dr with F = 3x2y2i + (2x3y − 1)j. From Exercise 1(b) F is a

conservative vector field. Find a scalar field φ so that ∇φ = F . Use property P4 to evaluate

the integral

C

F · dr where C is an integral with start-point (0, 0) and end point (1, 4).

3. For the following conservative vector fields F , find a scalar field φ such that ∇φ = F and

hence evaluate the I =

C

F · dr for the contours C indicated.

(a) F = (4x3y − 2x)i + (x4 − 2y)j; any path from (0, 0) to (2, 1).

(b) F = (ex + y3)i + (3xy2)j; closed path starting from any point on the circle x2 + y2 = 1.

(c) F = (y2 + sin z)i + 2xyj + x cos zk; any path from (1, 1, 0) to (2, 0, π).

(d) F =1

xi + 4y3z2j + 2y4zk; any path from (1, 1, 1) to (1, 2, 3).

Answers

1. (a) No, (b) Yes, (c) No, (d) Yes

2. x3y2 − y + C, 12

3. (a) x4y − x2 − y2, 11; (b) ex + xy3, 0; (c) xy2 + x sin z, −1; (d) ln x + y4z2,143

HELM (2008):


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4. Vector line integrals

It is also possible to form less commonly used integrals of the types:

C

f(x, y, z) dr and

C

F (x, y, z)× dr.

Each of these integrals evaluates to a vector.

Remembering that dr = dx i + dy j + dz k, an integral of the form

C

f(x, y, z) dr becomes

C

f(x, y, z)dx i +

C

f(x, y, z) dy j +

C

f(x, y, z)dz k. The first term can be evaluated by

expressing y and z in terms of x. Similarly the second and third terms can be evaluated by expressingall terms as functions of y and z respectively. Alternatively, all variables can be expressed in termsof a parameter t. If an integral is two-dimensional, the term in z will be absent.

Example 13


C

xy2dr where C represents the contour y = x2 from (0, 0)

to (1, 1).

Solution

This is a two-dimensional integral so the term in z will be absent.

I =

C

xy2dr

=

C

xy2(dxi + dyj)

=

C

xy2dx i +

C

xy2

dy j

=

1

x=0

x(x2)2dx i +

1

y=0

y1/2

y2

dy j

=

1

0

x5dx i +

1

0

y5/2

dy j

=

1

6x

6

1

0

i +

2

7x

7/2

1

0

j

=1

6i +

2

7j

26 HELM (2008):


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Example 14

Find I =

C

xdr for the contour C given parametrically by x = cos t, y = sin t,

z = t − π starting at t = 0 and going to t = 2π, i.e. the contour starts at(1, 0,−π) and finishes at (1, 0, π).

Solution

The integral becomes

C

x(dx i + dy j + dz k).

Now, x = cos t, y = sin t, z = t− π so dx = − sin t dt, dy = cos t dt and dz = dt. So

I =

2π

0

cos t(− sin t dt i + cos t dt j + dt k)

= − 2π

0

cos t sin t dt i +

2π

0

cos2t dt j +

2π

0

cos t dt k

= −1

2

2π

0

sin 2t dt i +1

2

2π

0

(1 + cos 2t) dt j +

sin t

2π

0k

=1

4

cos 2t

2π

0i +

1

2

t +

1

2sin 2t

2π

0

j + 0k

= 0i + π j = πj

Integrals of the form

C

F × dr can be evaluated as follows. If the vector field F = F1i+F2j +F3k

and dr = dx i + dy j + dz k then:

F × dr =

i j k

F1 F2 F3

dx dy dz

= (F2 dz − F3 dy)i + (F3 dx− F1 dz)j + (F1 dy − F2 dx)k

= (F3j − F2k)dx + (F1k − F3i)dy + (F2i− F1j)dz

There are thus a maximum of six terms involved in one such integral; the exact details may dictatewhich method to use.

HELM (2008):


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Example 15


C

(x2i + 3xyj)× dr where C represents the curve y = 2x2

from (0, 0) to (1, 2).

Solution

Note that the z components of both F and dr are zero.

F × dr =

i j k

x2 3xy 0

dx dy 0

= (x2dy − 3xydx)k and

C

(x2i + 3xyj)× dr =

C

(x2dy − 3xydx)k

Now, on C, y = 2x2 dy = 4xdx and

C

(x2i + 3xyj)× dr =

C

x2dy − 3xydxk

=

1

x=0

x

2 × 4xdx− 3x× 2x2dx

k

=

1

0

−2x3dxk

= −1

2x

4

1

0

k

= −1

2k

28 HELM (2008):


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Force on a loop due to a magnetic field

Introduction

A current I in a magnetic field B is subject to a force F given by

F = I dr ×B

where the current can be regarded as having magnitude I and flowing (positive charge) in thedirection given by the vector dr. The force is known as the Lorentz force and is responsible for theworkings of an electric motor. If current flows around a loop, the total force on the loop is given bythe integral of F around the loop, i.e.

F =

(I dr ×B) = −I

(B × dr)

where the closed path of the integral represents one circuit of the loop.

Figure 5: The magnetic field through a loop of current

Problem in words

A current of 1 amp flows around a circuit in the shape of the unit circle in the Oxy plane. A magneticfield of 1 tesla (T) in the positive z-direction is present. Find the total force on the circuit loop.


Choose an origin at the centre of the circuit and use polar coordinates to describe the position ofany point on the circuit and the length of a small element.

Calculate the line integral around the circuit to give the force required using the given values ofcurrent and magnetic field.


The circuit is described parametrically by

x = cos θ y = sin θ z = 0

with

dr = − sin θ dθ i + cos θ dθ j

B = B k

HELM (2008):


29

since B is constant. Therefore, the force on the circuit is given by

F = −IB

k × dr = −

k × dr (since I = 1 A and B = 1 T)

where

k × dr =

i j k

0 0 1

− sin θ dθ cos θ dθ 0

=− cos θ i− sin θ j

dθ

So

F = − 2π

θ=0

− cos θ i− sin θ j

dθ

=sin θ i− cos θ j

2π

θ=0

= (0− 0) i− (1− 1) j = 0

Hence there is no net force on the loop.

Interpretation

At any given point of the circle, the force on the point opposite is of the same magnitude but oppositedirection, and so cancels, leaving a zero net force.

Tip: Use symmetry arguments to avoid detailed calculations whenever possible!

30 HELM (2008):


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A scalar or vector involved in a vector line integral may itself be a vector derivative as this nextExample illustrates.

Example 16

Find the vector line integral

C

(∇ ·F ) dr where F is the vector x2i+2xyj +2xzk

and C is the curve y = x2, z = x3 from x = 0 to x = 1 i.e. from (0, 0, 0) to(1, 1, 1). Here ∇ · F is the (scalar) divergence of the vector F .

Solution

As F = x2i + 2xyj + 2xzk, ∇ · F = 2x + 2x + 2x = 6x.The integral

C

(∇ · F ) dr =

C

6x(dx i + dy j + dz k)

=

C

6x dx i +

C

6x dy j +

C

6x dz k

The first term is

C

6x dx i =

1

x=0

6x dx i =

3x2

1

0

i = 3i

In the second term, as y = x2 on C, dy may be replaced by 2x dx so

C

6x dy j =

1

x=0

6x× 2x dx j =

1

0

12x2dx j =

4x3

1

0

j = 4j

In the third term, as z = x3 on C, dz may be replaced by 3x2 dx so

C

6x dz k =

1

x=0

6x× 3x2dx k =

1

0

18x3dx k =

9

2x

4

1

0

k =9

2k

On summing,

C

(∇ · F ) dr = 3i + 4j +9

2k.

Task

Find the vector line integral

C

fdr where f = x2 and C is

(a) the curve y = x1/2 from (0, 0) to (9, 3).

(b) the line y = x/3 from (0, 0) to (9, 3).

HELM (2008):


31

Your solution

Answer

(a)

9

0

(x2i +

1

2x

3/2j)dx = 243i +

243

5j, (b)

9

0

(x2i +

1

3x

2j)dx = 243i + 81j.

Task

Evaluate the vector line integral

C

F × dr when C represents the contour

y = 4−4x, z = 2−2x from (0, 4, 2) to (1, 0, 0) and F is the vector field (x−z)j.

Your solution

Answer 1

0

(4− 6x)i + (2− 3x)k = i +1

2k

32 HELM (2008):


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Exercises

1. Evaluate the vector line integral

C

(∇ · F ) dr in the case where F = xi + xyj + xy2k and

C is the contour described by x = 2t, y = t2, z = 1 − t for t starting at t = 0 and going tot = 1.

2. When C is the contour y = x3, z = 0, from (0, 0, 0) to (1, 1, 0), evaluate the vector lineintegrals

(a)

C

∇(xy)× dr

(b)

C

∇× (x2

i + y2k)

× dr

Answers

1.

C

(1 + x)(dx i + dy j + dz k) = 4i +7

3j − 2k,

2. (a) k

C

y dy − x dx = 0k = 0, (b) k

C

2y dy = 1k = k

HELM (2008):


33

Index for Workbook 29

Area theorem 78 Conservat ive vec tor f ie ld 13, 19

Current cont inu ity equation 47

Current in l ine 69

Current in loop 29

Cyl inder 70

Cyl indr ical polar coordinates 46, 59

Divergence theorem – See Gauss

Magnet ic F ie ld 29

Magnet ic F lux 44

Ohm’s law 48

Py thagoras ’ theorem 37

Scalar l ine integral 8

Scalar product 8

Stokes ’ theorem 56-62

Surface in tegra ls 35-50

Electr ic cur rent 47

Electr ic f ie ld 21, 65, 69, 70

Electr ic motor 29

Fie ld s trength 69, 70 Gauss ’ l aw 65, 67

Gauss ’ theorem 63-72

Green’s theorem 73-80

Integral vec tor theorems

- Gauss ’ 63-72 - Green’s 73-80

- S tokes ’ 56-62 Integra ting

- scalar f ield 37 - vec tor f ie ld 42

L igh tn ing s tr ike 47

Line in tegra ls

- scalar produc ts 8

- vec tor 26

Lorentz force 29

Un it normal 35 Volume in tegra ls 50-54 Work 21 EXERCISES 18, 25, 33, 42, 50, 52, 54, 62, 72, 80 ENGINEERING EXAMPLES 1 Work done mov ing a charge in an

electr ic f ield 21

2 Force on a loop due to a magnetic f ield 29

3 Magnet ic f lux 44

4 The cur rent cont inu ity equat ion - 47

5 Gauss ’ l aw 65

6 F ie ld strength around a charged l ine 69

7 Field s trength on a cy l inder 70

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About the HELM Projectqub.ac.uk/helm/HELM_2008/pages/...Vector_Calculus.pdf · Integral Vector Calculus 29.1 Line Integrals 2 29.2 Surface and Volume Integrals 34 29.3 Integral Vector